problem solving involving angle of elevation

Solved Example Problems | Trigonometry | Mathematics - Problems involving Angle of Elevation | 10th Mathematics : UNIT 6 : Trigonometry

Chapter: 10th mathematics : unit 6 : trigonometry, problems involving angle of elevation.

Heights and Distances

In this section, we will see how trigonometry is used for finding the heights and distances of various objects without actually measuring them. For example, the height of a tower, mountain, building or tree, distance of a ship from a light house, width of a river, etc. can be determined by using knowledge of trigonometry. The process of finding Heights and Distances is the best example of applying trigonometry in real-life situations. We would explain these applications through some examples. Before studying methods to find heights and distances, we should understand some basic definitions.

Line of Sight

The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.

Theodolite is an instrument which is used in measuring the angle between an object and the eye of the observer. A theodolite consists of two graduated wheels placed at right angles to each other and a telescope. The wheels are used for the measurement of horizontal and vertical angles. The angle to the desired point is measured by positioning the telescope towards that point. The angle can be read on the telescope scale.

Angle of Elevation

The angle of elevation is an angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level. That is, the case when we raise our head to look at the object. (see Fig. 6.7)

Angle of Depression

The angle of depression is an angle formed by the line of sight with the horizontal when the point is below the horizontal level. That is, the case when we lower our head to look at the point being viewed. (see Fig. 6.8)

The angle of elevation and depression are usually measured by a device called clinometer.

·         From a given point, when height of a object increases the angle of elevation increases.

 If h 1 > h 2 then α > β

·         The angle of elevation increases as we move towards the foot of the vertical object like tower or building.

If d 2 < d 2 then β > α

In this section, we try to solve problems when Angle of elevation are given.

Example 6.18

Calculate the size of ∠ BAC in the given triangles.

(i) In right triangle ABC [see Fig.6.12(a)]

 tan θ = opposite side / adjacent side = 4/5

 θ  = 38.7° (since tan 38.7° = 0.8011)

∠ BAC = 38.7°

(ii) In right triangle ABC [see Fig.6.12(b)]

tan θ = 8/3

θ  = 69.4° (since tan 69.4° = 2.6604)

∠ BAC = 69.4°

Example 6.19

A tower stands vertically on the ground. From a point on the ground, which is 48 m away from the foot of the tower, the angle of elevation of the top of the tower is 30° . Find the height of the tower.

Let PQ be the height of the tower.

Take PQ = h   and QR is the distance between the tower and the point R . In right triangle PQR , ∠ PRQ = 30°

Therefore the height of the tower is 16√3 m

Example 6.20

A kite is flying at a height of 75 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60° . Find the length of the string, assuming that there is no slack in the string.

Let AB be the height of the kite above the ground. Then, AB = 75.

Let AC be the length of the string.

In right triangle ABC , ∠ ACB = 60°

sin θ = AB/AC

sin 60° = 75/ AC

 gives √3/2 = 75/ AC so AC = 150/√3 = 50√3 m

Hence, the length of the string is 50√3 m.

Example 6.21

Two ships are sailing in the sea on either sides of a lighthouse. The angle of elevation of the top of the lighthouse as observed from the ships are 30° and 45° respectively. If the lighthouse is 200 m high, find the distance between the two ships. (√3=1.732)

Let AB be the lighthouse. Let C and D be the positions of the two ships.

Then, AB = 200 m.

∠ ACB = 30° , ∠ ADB = 45°

In right triangle BAC , tan 30° =  AB/AC

1/√3 = 200/AC gives AC = 200√3      ... (1)

In right triangle BAD , tan 45° = AB/AD

1 = 200/ AD     gives AD = 200     ... (2)

Now, CD = AC + AD  = 200√3 + 200 [by (1) and (2) ]

CD = 200(√3 + 1) = 200 × 2. 732 = 546.4

Distance between two ships is 546.4 m.

Example 6.22

From a point on the ground, the angles of elevation of the bottom and top of a tower fixed at the top of a 30 m high building are 45° and 60° respectively. Find the height of the tower.  (√3=1.732)

Solution 

Let AC be the height of the tower.

Let AB be the height of the building. Then, AC = h metres, AB = 30 m

In right triangle CBP , ∠ CPB = 60°

In right triangle ABP , ∠ APB = 45° 

h = − 30(√3 - 1) = 30 (1.732 – 1) = 30(0.732) = 21.96

Hence, the height of the tower is 21.96 m.

Example 6.23

A TV tower stands vertically on a bank of a canal. The tower is watched from a point on the other bank directly opposite to it. The angle of elevation of the top of the tower is 58 ° . From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30 ° . Find the height of the tower and the width of the canal. (tan 58 ° = 1.6003)

Let AB be the height of the TV tower.

Let BC be the width of the canal.

Hence, the height of the tower is 17.99 m and the width of the canal is 11.24 m.

Example 6.24

An aeroplane sets off from G on a bearing of 24 ° towards H , a point 250 km away. At H it changes course and heads towards J on a bearing of 55° and a distance of 180 km away.

(i) How far is H to the North of G ?

(ii) How far is H to the East of G ?

(iii) How far is J to the North of H ?

(iv) How far is J to the East of H ?

(i) In right triangle GOH , cos 24° = OG/GH

0.9135 = OG/ 250 ; OG = 228.38 km

Distance of H to the North of G = 228.38 km

(ii) In right triangle GOH ,

sin 24° = OH/GH

0. 4067 = OH/ 250 ; OH = 101. 68

Distance of H to the East of G = 101. 68 km

(iii) In right triangle HIJ ,

sin 11° = IJ/HJ

0. 1908 = IJ/ 180 ; IJ = 34. 34 km

Distance of J to the North of H = 34. 34 km

(iv) In right triangle HIJ ,

cos 11° = HI/HJ

0. 9816 = HI / 180 ; HI = 176. 69 km

Distance of J to the East of H = 176. 69 km

Example 6.25

Two trees are standing on flat ground. The angle of elevation of the top of both the trees from a point X on the ground is 40° . If the horizontal distance between X and the smaller tree is 8 m and the distance of the top of the two trees is 20 m, calculate

(i) the distance between the point X and the top of the smaller tree.

(ii) the horizontal distance between the two trees.  (cos 40° = 0. 7660)

Let AB be the height of the bigger tree and CD be the height of the smaller tree and X is the point on the ground.

(i) In right triangle XCD , cos 40°= CX/XD

XD = 8 / 0.7660   = 10. 44 m

Therefore the distance between X and top of the smaller tree = XD = 10.44 m

(ii) In right triangle XAB ,

Therefore the horizontal distance between two trees = AC = 15.32 m

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Angles of Elevation and Angles of Depression Word Problems

Angles of elevation and angles of depression word problems are some of the most common problems in studies of trigonometry. These types of problems are famous for requiring the use of trigonometric ratios to relate a given angle and side length to find an unknown side length.

Whether it is the base of the cliff or the height of the lamp post, understanding how to solve angles of elevation and angles of depression word problems is an essential problem solving skill! 

So let’s dig into it! It’s time to …  elevate …  your understanding!

What are Angles of Depression and Elevation?

Picture standing at the top of a tower and looking straight ahead along an imaginary horizontal line of sight. Suddenly, someone at ground level calls your name! You tilt your head downward from the imaginary horizontal line toward the ground through some angle. 

As it turns out, this angle is the  angle of depression!  This angle is sometimes referred to as the angle of declination since there is a “decline” between the horizontal line of sight and the object being observed. An angle of depression must always be an acute angle.

Now, imagine that you are the person shouting up toward the top of the tower. You begin by looking straight ahead at the base of the tower, and tilt your head from ground level up toward the top of the tower.

The angle that you are tilting your head through is the  angle of elevation.  This angle is sometimes referred to as the angle of inclination since there is an “incline” between the horizontal line of sight and the top of the object. An angle of elevation must also always be an acute angle.

There are many famous angle theorems that can help us solve problems involving triangles. However, if you think back to your understanding of alternate interior angles , you should be able to see that these two angles are actually equal! This is because the line of sight and the ground are parallel lines.

Since we have a “Z-pattern”, we know that the angle of depression is equal to the angle of elevation ! This is an important piece of knowledge that is going to help you solve any given angles of elevation or angles of depression word problems that you come across! 

Before we take a look at the first example, watch this short video for an excellent visual representation of angles of elevation and angles of depression. I promise you won’t forget the difference after watching this video!

Solving Angles of Elevation and Angles of Depression Word Problems

When it comes to solving angles of elevation and angles of depression word problems, the first step is to always start with a diagram to visualize the problem. Look for key phrases that give you hints about what specific angles and side lengths might be involved. Phrases such as  the height of the first building  give you an indication that that particular building will be important in your calculation.

From here, think about what trigonometric ratios will be involved. For example, the tangent ratio is used when you are relating the angle, the opposite, and the adjacent sides.

Let’s take a look at a few examples!

Example #1: Angles of Depression Word Problems

You are looking out a window at the very top of a tower. The height of the tower is known to be 50 feet. From the top of the building, you measure the angle of depression of a car to be 30 degrees. How far is the car from the building to the nearest foot?

To begin, we draw a diagram to help us visualize this scenario. The most important pieces of information given in this problem are the height of the tower (one of the sides of the right triangle) and the angle of depression of the car. Remember, since we are working with a building, we know that we have a right angle between the building and the ground.

The key trick to most angles of depression word problems is using your knowledge of alternate interior angles. This works since our line of sight and the ground are parallel lines!

We can use our understanding of alternate interior angles to show that the angle of depression from the top of the building is actually equal to the angle of elevation from the car up toward the tower!

Therefore, each of these angles are now known to be 30 degrees.

Looking at this new angle, we have the length of the side  opposite  to the angle (the height of the building) and we want to solve for length of the side  adjacent to the angle   (the distance from the car to the base of the tower). The trigonometric ratio relating the opposite and adjacent side lengths is the tangent ratio!

Setting up this ratio and solving for  x  results in:

$$ \begin{split} tan30^\circ &= \frac{50}{x} \\ \\ x &= \frac{50}{tan30^\circ} \\ \\ x &= 86.6 \\ \end{split} $$

Therefore, the car is 87 feet away from the base of the tower (rounded to the nearest foot).

Example #2: Angles of Elevation Word Problems

You are standing on the ground and see two buildings in front of you that are the same distance away from you. You measure the angle of elevation to the top of the first building to be 45 degrees, and the angle of elevation to the top of the second building to be 60 degrees. You know that the height of the first building is 40 feet. What is the height of the second building to the nearest foot?

Just like the last problem, we start by sketching a diagram to help visualize this scenario. This scenario is slightly more complex since we are working with two buildings, so a diagram is important!

As it stands, the problem has not given us enough information to solve for the height of the second building, h . In order to solve the problem, we first need to determine the horizontal distance, d , between you and the base of the buildings. Once we have determined this horizontal distance, we can use it to solve for the height of the second building!

We use the height of the first building and the 45 degree angle (the angle of elevation) to set up a tangent ratio:

$$ \begin{split} tan45^\circ &= \frac{40}{d} \\ \\ d &= \frac{40}{tan45^\circ} \\ \\ d &= 40 \\ \end{split} $$

So we know the horizontal distance, d , is also 40 feet. From here, we can use the horizontal distance and the 60 degree angle of elevation to find the height of the second building!

$$ \begin{split} tan60^\circ &= \frac{h}{40} \\ \\ 40 \times tan60^\circ &= h \\ \\ h &= 69.28 \\ \end{split} $$

Therefore, the height of the second building is 69 feet!

Review: Solving Angles of Elevation and Depression Word Problems

Each angle of elevation or depression problem that you encounter will have similarities. You will always have to use the information given and a vertical line formed by some object to find the height of the tree, building, or tower.

If you aren’t give one, drawing a diagram should always be your first step. This is important to help you visualize the angles of elevation or depression that you are given. And as you have seen, having a good understanding of alternate interior angles and trigonometric ratios is also important!

For more practice on solving these types of problems, check out this angles elevation and angle of depression word problems worksheet !

My hope is that these examples have helped you understand the difference between angles of elevation and angles of depression. With this knowledge and a little practice, you will be able to solve any related problem that you come across!

Did you find this overview of angles of elevation and depression helpful? Share this post and subscribe to Math By The Pixel on YouTube for more helpful mathematics content!

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• Angle Of Elevation

Angle of Elevation

The angle of elevation is a widely used concept related to height and distance, especially in trigonometry. It is defined as an angle between the horizontal plane and oblique line from the observer’s eye to some object above his eye. Eventually, this angle is formed above the surface. As the name itself suggests, the angle of elevation is so formed that it is above the observer’s eye.

For example, an observer is looking at a bird sitting at the rooftop, then there is an angle formed, which is inclined towards the bird from the observer’s eye. This elevation angle is used in finding distances, heights of buildings or towers, etc with the help of trigonometric ratios , such as sine, cosine and tangent.

Table of contents:

Horizontal Line

• Line of sight

Angle of Elevation Definition

The angle of elevation is an angle that is formed between the horizontal line and the line of sight. If the line of sight is upward from the horizontal line, then the angle formed is an angle of elevation.

In the above figure, you can see, an observer is looking at the object, standing on the ground, forming an angle θ with the line of sight and horizontal line. Here, if we join an imaginary line between the object and end of the horizontal line, a right angle triangle is formed. Thus we can use here trigonometry concept to find the distance of the observer from the tower or building. Height of the tower or building or the height at which the object is kept will be considered as perpendicular and the horizontal line will be considered as adjacent side of the triangle formed. The related terminology are given here.

Also, read:

Terms Used for Angle of Elevation

The three terms related to the angle of elevation are

If two rays or two line segments meet at a common endpoint, then the point is known as the vertex. Two straight lines meet at a common point is said to form an angle. Angle play an important role, and some other definitions of angle are:

• The angle is a gap in between two lines which connect on one side.
• The angles are measured in degrees.

A straight line on the coordinate flat surface where all points on the line have the same y-coordinate. The angle and horizontal line combine to form the angle of elevation.

Line of Sight

The line which is drawn from the eyes of the observer to the point being viewed on the object is known as the line of sight.

Here, the object is kept above the line of sight of the observer. If we know the elevation angle, then we can easily determine the distance or the altitude. The reason why we use here trigonometric functions is that the angle formed with the respect to the observer’s eye to the top of a building or a tower produces an imaginary right triangle, where the height of the building or tower is considered as the perpendicular of the triangle.

Angle of Elevation Formula

The formula for finding the angle of elevation depends on knowing the information such as the measures of the opposite, hypotenuse, and adjacent side to the right angle. If the distance from the object and height of the object is given, then the formula for the angle of elevation is given by

Tangent of the angle of elevation = Height of the Object / Distance from the object

Tan θ = Opposite Side/Adjacent Side

Angle of Elevation and Depression Comparison

The angle of depression is just the opposite scenario of the angle of elevation. In this case, the observer is standing at height and the object is kept below the line of sight of the observer. We can define it as if the object is kept below the eye level of the observer, then the angles formed between the horizontal line and the observer’s line of sight is called the angle of depression .

The formula of the angle formed here is given by;

tangent of angle of depression = Opposite side/adjacent side

Angle of Elevation Examples

Question 1: Find the value of θ in the given figure.

In the given triangle ABC, AC = 335 ft, BC = 249 ft

To find ∠A = θ

tan θ = Opposite side/ Adjacent side

tan θ = BC / AC

tan θ = 249/ 335

tan θ = 0.74

Therefore, θ = tan -1 (0.74) = 36

Hence, the value of θ = 36 degrees.

Question 2: If the angle of inclination formed by an observer to an object, kept above a tower, is 45 degrees. Find the horizontal distance between the observer and the height of the tower is 150 ft.

Solution: Given, θ = 45

Height of tower = 150 ft.

To find: the distance between the observer and base of the tower.

By the formula, we know,

tan θ  = Height of the tower/Distance between observer and tower

tan 45 = 150/D

since tan 45 = 1

D = 150 ft.

Hence, the distance between the observer and tower is equal to the height of the tower.

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Angle of Elevation

The angle of elevation is the angle between the horizontal line and the line of sight which is above the horizontal line. It is formed when an observer looks upwards. Suppose you are standing at the terrace of a building and looking upwards at the sky or at the sun or moon. The angle thus formed between your height from the ground level and the line of sight formed is called the angle of elevation.

Angle of Elevation Definition

The angle of elevation in math is "the angle formed between the horizontal line and the line of sight when an observer looks upwards is known as an angle of elevation". It is always at a height that is greater than the height of the observer. The opposite of the angle of elevation is the angle of depression which is formed when an observer looks downwards. It is important to learn about the angle of elevation and depression while studying heights and distances in trigonometry . The three general words associated with the angle of elevation are angles, horizontal lines, and line of sight.

Let the observer at 'O' is observing an object that is at 'A'. Then the horizontal line is OB and the line of sight is OA. Then the angle formed between OA and OB which is angle AOB is the angle of elevation.

Angle of Elevation Formula

The angle of elevation formula is no different from the formulae of trigonometric ratios . With the help of the formulae given below, we can find the angle of elevation depending on which two sides of the triangle are known. For example, if we have to find the angle of elevation when the height of the object from the horizontal line and the length of the line of sight are known, we can use the sine formula .

For example, to calculate the angle of elevation for an object at the distance of 10 units from the horizontal line (y=10) and 12 units from the observer w.r.t. the horizontal line (x=12), we write, tan θ = 10/12, which can be reduced to tan θ = 5/6. Therefore, the value of θ obtained is tan -1 (5/6). This is the required angle of elevation.

Angle of Elevation Vs Angle of Depression

The angle of elevation and angle of depression are opposites of each other. In an angle of elevation, the object is placed above the observer, while in the case of the angle of depression, it is placed below the observer. If you are standing at your terrace and looking at the sun, then the angle of elevation will be formed. On the other hand, if you will look at the dog standing on the road from your terrace, then the angle of depression will be formed. In both cases, we use trigonometry angles to find the heights and distances. Let us understand the difference between the angle of elevation and depression from the table given below.

In the above diagram, θ is the angle of elevation and α is the angle of depression.

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Angle of Elevation Examples

Example 1: If a girl is standing at point P, which is 8 units away from a building, making an angle of elevation of 45° with point Q, find the height of the building.

Solution: Given that PR=8 units, and ∠QPR=45°. To find the height of the building (QR), we can use the angle of elevation formula tan θ=QR/PR.

tan 45° = QR/8

We know that tan 45° is 1, so,

Answer: Therefore, the height of the building is 8 units.

Example 2: Find the value of x in the given figure.

Solution: In this figure, there are two angles of elevation given, one is 30° and the other one is 45°. In △POQ, ∠PQO = 30 degrees and OQ=27 feet. Apply the angle of elevation formula tan θ = PO/OQ, we get tan 30 = h/27. The value of tan 30 is 1/√3.

1/√3 = h/27

h = 3 √3/√3

Now, apply the same formula in △POR, we get tanθ = PO/OR.

tan 45 = 3/x

The value of tan 45 is 1, and PO = 3 ft.

Answer: Therefore, the value of x is 3 ft.

Example 3: Ryan is flying a kite that makes an angle of elevation of 30° with the ground. Determine how

high the kite is above the ground when she has let out 100 m of string.

Let 'h' be the height of the kite from the ground. Then

Applying sine for the figure,

sin 30 = h/100

We know that sin 30 = 1/2.

1/2 = h/100

h = 100/2 = 50 m

Answer: The kite is 50 m above the ground.

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Practice Questions on Angle of Elevation

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FAQS on Angle of Elevation

What is meant by angle of elevation.

An angle formed when an observer looks at an object placed above its height with respect to the eye level or the horizontal line is known as the angle of elevation . For example, the angle formed between the line of sight and the horizontal line when the sun is observed by a man on the earth is an angle of elevation.

How do you Find the Angle of Elevation?

The angle of elevation can be found when we are given with any two sides of the right triangle formed between the observer and the object. We can use the trigonometric formulae to find the angle of elevation.

How are Angle of Elevation and Depression Related?

The angle of elevation and angle of depression are both measured with respect to the horizontal axis or the horizontal line. The only difference is that when an observer looks upwards at an object, then the angle of elevation is formed, while when she/he looks downwards at an object, the angle of depression is formed.

What is the Angle of Elevation of the Sun?

The angle of elevation of the sun is the angle formed between the horizontal line and your line of sight when you look at the sun. It will keep on changing because the position of the sun keeps on changing. So, even if you stand at the same point in the morning and in the afternoon, the angle of elevation will differ.

Can an Angle of Elevation be More Than 90?

No, an angle of elevation cannot be more than 90 degrees . It always forms a right-angled triangle with the object and the horizontal line. In the right triangle, one angle is 90 degrees which is the angle opposite to the line of sight. So, it is obvious that the other two angles will be less than 90 degrees in order to satisfy the angle sum property of a triangle. Therefore, an angle of elevation cannot be more than 90 degrees.

What is the Formula for Angle of Elevation?

There are three formulae that can be used to find the angle of elevation. The angle of elevation formulas are given below:

• sin θ = perpendicular/hypotenuse
• cos θ = base / hypotenuse
• tan θ = perpendicular/base

How to Find Height with Angle of Elevation?

To find the height with the angle of elevation, we need to use the trigonometric formulas mentioned above based on which two sides of the right triangle are given.

How Do You Find Angle of Elevation and Depression?

Angle of elevation is the angle between a horizontal line of sight and the object when a person is looking up at an object. Whereas the angle of depression is the angle between the horizontal line of sight and the object when a person is looking down at an object.

What is the Importance of Angle of Elevation and Depression?

One of the main aspects of using the angle of elevation and depression is that it is used mostly in word problems in trigonometry when there is a line of sight involved. These angles are used in solving trigonometric problems such as sine, cosine, and tangent along with inverse trigonometric functions .

What is the Altitude of Sun in Angle of Elevation Problems?

The altitude of the sun means the angle of elevation of the sun from the observer. For finding this, we need to use sin, cos, or tan according to the given information.

Helping with Math

Angle of Elevation

Introduction.

In mathematics and physics, trigonometry has its applications. Mathematical concepts like height, distances, angle of elevation, and angle of depression are examples of applications of trigonometry . 

Let us imagine; for example, an observer is looking at an aircraft—an angle of elevation formed towards the aircraft from the observer’s view. The angle of elevation helps us find distances using trigonometric ratios like sine, cosine, and tangent.

This article will discuss the angle of elevation, the related terms, and how to calculate the angle of elevation.

The angle generated by the line from an object to the observer’s eye and the horizontal line or plane is known as the angle of elevation. The line of sight is the line the observer’s eye is on. As the word says, the angle of elevation is thus formed so that it is above the observer’s eye.

For example, an observer is looking at an object, as shown in the figure below. The angle of elevation is represented by θ. 

Here, a right-angle triangle is created if we connect the object vertically and the horizontal line. Thus, we may utilize trigonometry to compute the observer’s distance from the object. The horizontal line will be regarded as the adjacent side of the formed triangle, and the height of the object will be looked at as the opposite side of θ.

Terms Related to Angle of Elevation

These are the terms we must be familiar with when dealing with the angle of elevation.

When two straight rays or lines meet at a single point, an angle is created. Degrees are used to express angles .

Line of Sight

The line of sight is the line drawn from the observer’s eye to the object being viewed. It is an oblique line which is therefore not horizontal nor vertical. The line of sight, together with the horizontal line, forms the angle of elevation. 

Horizontal Line

The horizontal line is a straight line on a flat surface. In a coordinate system, a horizontal line has points with the same y-coordinate. The line of sight, together with the horizontal line, form the angle of elevation. 

Observer’s Eye

The observer’s eye is the position where the line of sight and the horizontal line meet.

The Angle of Elevation Formulas

There are formulas that will help us calculate the angle of elevation. 

Let us refer to the figure below to show how we use the formulas. Recall that a right-angle triangle is formed if an imaginary line is drawn connecting the object and the horizontal line’s end. As a result, we may use the trigonometry concept to determine how distant the observer is from the object. The horizontal line will be regarded as the adjacent side of the constructed triangle , and the height of the object will be looked at as the opposite side of θ. 

In the figure, side a is the opposite side of , side b is the adjacent side, and side c is the length of the line of sight.

To determine the angle of elevation, we can use the formulas below, provided we know the angles’ two sides. 

If we must calculate the angle of elevation given the line of sight and the height of the object from the horizontal line, we can use the sine formula θ = ($\frac{a}{c}$). 

If we must calculate the angle of elevation given the line of sight and the adjacent side of , we can use the cosine formula θ = ($\frac{b}{c}$).

If we must calculate the angle of elevation given the opposite side and the adjacent side of , we can use the tangent formula θ = ($\frac{a}{b}$).

The Angle of Elevation vs. Angle of Depression

Angles of elevation and depression are opposites of one another. The object is positioned above the observer at an angle of elevation, whereas it is positioned below the observer at an angle of depression. The angle of elevation is generated if you are standing and looking at an object on the rooftop. Whereas, if you were to look at the plants on the ground, an angle of depression is formed. In both cases, we can use trigonometry concepts to find the heights and distances. 

The table below shows the difference between the angle of elevation and depression.

Find the value of the angle of elevation ( θ ) in each of the given figures.

( a ) In the given triangle ABC, ∠A is the angle of elevation, side AC is the opposite side, while side BC is the adjacent side of the angle θ. Since we know two sides of the triangle, opposite and the adjacent sides of the angle θ, we must use the formula for the tangent. 

AC (opposite side ) = 240 ft

BC ( adjacent side ) = 330 ft

Thus, we have,

tan tanθ=$\frac{opposite\: side}{adjacent\: side}$  

tan tanθ=$\frac{240 ft}{330 ft}$  

tan tanθ=0.72  

Therefore, θ=( 0.72 )

Hence, the value of θ ≈ 36°.

( b ) In the given triangle PRQ, ∠R is the angle of elevation, side PR is the length of the line of sight or the hypotenuse of the triangle, while side QR is the adjacent side of the angle θ. Since we know the lengths of the hypotenuse and the adjacent side of the angle θ, we must use the formula for cosine. 

PR (hypotenuse side ) = 26 m

QR ( adjacent side ) = 10 m

cos cosθ=$\frac{adjacent\: side}{hypotenuse\: side}$  

cos cosθ=$\frac{10 m}{26 m}$  

cos cosθ=0.38  

Therefore, θ=( 0.38 )

Hence, the value of θ ≈ 68°.

( c ) In the given triangle LMN, ∠M is the angle of elevation, side LM is the length of the line of sight or the hypotenuse of the triangle, while side LN is the opposite side of the angle θ. Since we know two sides of the triangle, hypotenuse and the opposite side of the angle θ, we must use the formula for sine. 

LM (hypotenuse side ) = 13 ft

LN ( opposite side ) = 5 ft

sin sinθ=$\frac{opposite\: side}{hypotenuse\: side}$  

sin sinθ=$\frac{5 ft}{13 ft}$  

sin sinθ=0.38  

Hence, the value of θ ≈ 22°.

( d ) In the given triangle DEF, ∠E is the angle of elevation, side DF is the opposite side of , while side EF is the adjacent side of the angle θ. Since we know two sides of the triangle, the adjacent and the opposite side of the angle θ, we must use the formula for the tangent. 

DF ( opposite side ) = 5 m

EF ( adjacent side ) = 12 m

tan tanθ=$\frac{5 m}{12 m}$  

tan tanθ≈0.42  

Therefore, θ=( 0.42 )

Hence, the value of θ ≈ 23°.

The shadow of a tree is 16 meters in length when the angle of elevation of the sun is 47°. Find the height of the tree.

The angle of elevation θ is 47°, and the length of the shadow, which is the adjacent side is 16 m. The opposite side of angle θ is the height of the tree. Let us use the formula for tangent since the problem involves the opposite and adjacent sides of the triangle. Thus, we have,

tan tan47°=$\frac{height\: of\: the\: tree}{16\: m}$  

tan 47° ( 16 m ) = height of the tree

17.16 m ≈ height of the tree

height of the tree ≈ 17.16°

Hence, the height of the tree is approximately 17.16 m.

At point M, a man is positioned 10 meters from a building, creating a 30° elevation with point T. Find the building’s height.

In the given scenario, the angle of elevation is 30° while the distance of the man horizontally from the base of the building is 10 meters. The opposite side of angle θ is the height of the building. Let us use the formula for tangent since the problem involves the opposite and adjacent sides of the triangle. Thus, we have,

tan tanθ=$\frac{opposite\: side}{adjacent\: side}$

tan tan30°=$\frac{height\: of\: the\: building}{10\: m}$  

tan 30° ( 10 m ) = height of the building

5.77 m ≈ height of the building

height of the building ≈ 5.77 m

Hence, the height of the building is approximately 5.77 m.

Marivic is flying a kite that is 60 degrees above the ground. Find the height of the kite above the ground after she has released 40 meters of string.

The angle of elevation θ is 60°. The length of the released string, 40 m, will be used as the hypotenuse of the right triangle. The opposite side of angle θ is the height of the kite above the ground. Let us use the formula for sine since the problem involves the opposite and hypotenuse of the triangle. Thus, we have,

sin sin60° = ( height of the kite from the ground ) / (40 m ) 

sin 60° ( 40 m ) = height of the kite from the ground 

20 √3 m = height of the kite from the ground 

height of the kite from the ground ≈ 34.64 m

Thus, the height of the kite from the ground is approximately 34.64 m.

A ladder climbs to the top of a vertical wall and makes a 45-degree angle with the ground. Determine the length of the ladder if the ladder’s foot is 5 meters from the wall.

The angle of elevation θ is 45°, while the ladder length is the right triangle’s hypotenuse. The distance between the foot of the wall is 5 meters, the distance of the adjacent side of the right triangle. Let us use the formula for cosine since the problem involves the adjacent and the hypotenuse of the triangle. Thus, we have,

cos cos45°=$\frac{adjacent\: side}{length\: of\: the\: ladder}$  

Length of the ladder = $\frac{5\: m}{cos\: cos45°}$

Length of the ladder = 5 √2 m

Length of the ladder ≈ 7 m

Hence, the length of the ladder is approximately 7 m.

The angle generated by the line from an object to the observer’s eye and the horizontal line or plane is known as the angle of elevation of the object as sighted by the observer. The line of sight is the line that the observer’s eye is on. As the word says, the angle of elevation is thus formed so that it is above the observer’s eye.

Frequently Asked Questions on Angle of Elevation ( FAQs )

What is meant by angle of elevation.

The angle generated by the line from an object to the observer’s eye and the horizontal line or plane is known as the angle of elevation of the object as sighted by the observer. The line of sight is the line that the observer’s eye is on. The angle of elevation is thus created, as the word “elevation” suggests, so it is above the observer’s eye.

What differentiates the angle of elevation from the angle of depression?

Angles of elevation and depression are opposites of one another. The object is positioned above the observer at an angle of elevation, whereas it is positioned below the observer at an angle of depression. The angle of elevation is generated if you are standing and looking at an object on the rooftop. On the other side, if you were to gaze at the plants on the ground, an angle of depression would be generated.

The table below shows the essential difference between the angle of elevation and the angle of depression.

In a triangle, what is the angle of elevation?

In terms of angle of elevation, the right-angle triangle is formed if an imaginary line is drawn connecting the object and the horizontal line’s end. 

As a result, we may use the trigonometry concept to determine how distant the observer is from the object. The horizontal line will be regarded as the adjacent side of the constructed triangle, and the height of the object will be looked at as the opposite side of θ. 

How do we apply trigonometric ratios to get the angle of elevation?

If we must calculate the angle of elevation given the adjacent side of , we can use the tangent formula θ = ($\frac{a}{b}$).

What are the terms related to the angle of elevation?

When two straight rays or lines meet at a single point, an angle is created. Degrees are used to express angles.

The line of sight is the line drawn from the observer’s eye to the object being viewed. It is an oblique line which is therefore not horizontal nor vertical. The line of sight, together with the horizontal line, form the angle of elevation. 

The horizontal line is a straight line on a flat surface. In a coordinate system, a horizontal line has points with the same y-coordinate . The line of sight, together with the horizontal line, form the angle of elevation. 

What is the angle of elevation, and how do you calculate it?

To find the angle of elevation, we can use the formulas below, provided we know the angles’ two sides. 

If we must calculate the angle of elevation given the opposite and the adjacent side of , we can use the tangent formula θ = ($\frac{a}{b}$).

For example, let us use the triangle LMN below.

In the given triangle LMN, ∠M is the angle of elevation, side LM is the length of the line of sight or the hypotenuse of the triangle, while side LN is the opposite side of the angle θ. Since we know the two sides of the triangle, the hypotenuse and the opposite side of the angle θ, we must use the formula for sine. 

sin sinθ=$\frac{5\: ft}{13\: ft}$  

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How Do You Solve a Problem Using an Angle of Elevation?

This tutorial provides a great real world application of math. You'll see how to use the tangent ratio to find the height of a hill. Take a look!

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• Terms of Use

ANGLE OF ELEVATION PRACTICE PROBLEMS

The angle of elevation is an angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level. That is, the case when we raise our head to look at the object.

In order to solve word problems, first draw the picture to represent the given situation.

Now, decide what we have to find from the given picture.

Mark the sides as opposite, hypotenuse and adjacent based on theta.

Now we have to choose a trigonometric ratio sin θ , cos  θ or tan θ based on the information that we have and the thing we have to find.

For example, if we have opposite side and we have to find the length of hypotenuse then we have to choose sin  θ. 

Problem 1 :

Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 1 0 √ 3 m.

BC - height of the tower

AC = hypotenuse side, BC = opposite side, AB = Adjacent side

Here we have to find  θ,  known sides are opposite and adjacent. Based on this information, we have to use tan  θ.

tan  θ  =  Opposite side / Adjacent side

tan  θ  =  BC/AB

  tan  θ  =  10 √3 /30

  tan  θ   =  √3/3

  tan  θ   = 1/ √3

θ  =  tan -1 ( 1/ √3)

  θ  =  tan -1 ( 1/ √3)

θ  =   30° or  π/6

Problem 2 :

A road is flanked on either side by continuous rows of houses of height 4 √ 3 m with no  space in between them. A pedestrian is standing on the median of the road facing a row  house. The angle of elevation from the pedestrian to the top of the house is 30° . Find the  width of the road.

AB = opposite side, BC = Adjacent side, AC = hypotenuse side

tan  θ  =  Opposite side / Adjacent side

tan 30  =  AB/BC

1/ √3  =  4√3/Distance from median of the road to house

BC  =  4√3(√3)  =  12

Width of the road :

=  2(BC)

=  2(12)

=  24 m

Problem 3 :

A tower stands vertically on the ground. From a point on the ground, which is 48 m away from the foot of the  tower, the angle of elevation of the top of the tower is 30° . Find the height of the tower.

In the triangle PRQ,

Opposite side = h, hypotenuse = PR and adjacent side = PQ

tan  θ = Opposite side / adjacent side

tan 30 = PQ / RQ

1/ √3 = h/48

h = 48 / √3

Rationalizing the denominator, we get

h = 48 √3/3

Then, height of the tower is 16√3 m.

Problem 4 :

A kite is flying at a height of 75 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60° . Find the length of the string, assuming that there is no slack in the string.

Length of string = hypotenuse

In triangle ABC, AB = opposite side

BC = Adjacent side

AC = Hypotenuse

sin  θ = Opposite side / Hypotenuse

sin 60 = AB / AC

√3/2 = 75/AC

AC = 75(2) /  √3

AC = 150 /  √3

So, the length of the string is 50√3 m.

Problem 5 :

Two ships are sailing in the sea on either sides of a lighthouse. The angle of elevation of the top of the lighthouse as observed from the ships are 30° and 45° respectively. If the lighthouse is 200 m high, find the distance between the two ships. ( √ 3 = 1.732 )

In triangles ABC and ABD.

tan 30 = AB / AC

1/ √3 = 200/AC

AC = 200√3 --(1)

tan 45 = AB / AD

1  = 200/AD

AD = 200 --(2)

Distance between two ships = AC + AD

= 200 √3 + 200

= 200(√3 + 1)

= 200(1.732 + 1)

= 200(2.732)

So, the distance between two ships is 546.4 m.

Problem 6 :

From a point on the ground, the angles of elevation of the bottom and top of a tower fixed at the top of a 30 m high building are 45° and 60° respectively. Find the height of the tower. ( √ 3 = 1.732)

In triangle ABP,

tan 45 = AB / BP

1 = 300 / BP

BP = 300 ----(1)

tan 60 = (AB + AC) / BP

√ 3  = (300 + h) / BP

BP = (300 + h)/ √ 3 ----(2)

300 = (300 + h) /  √ 3

300  √ 3 = 300 + h

h = 300 √ 3 - 300

h = 300( √ 3 - 1)

= 300(1.732 - 1)

= 300 (0.732)

Height of the tower = 300 + 219.6

So, the height of the tower is 519.6 m.

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Angles of Elevation and Depression: Examples

Related Topics: More Lessons for Grade 9 Math Math Worksheets

Examples, solutions, videos, worksheets, and activities to help students learn how to find the angle of elevation and depression using trigonometry.

The following figures show some examples of angle of elevation and angle of depression. Scroll down the page for more examples and solutions on solving problems with angle of elevation or angle of depression.

Angles of Depression and Elevation Learn about angles of depression and elevation.

Find the height of a dam using angle of elevation and the height of a helicopter using the concept of angle of depression Examples:

• From a point 340 m from the base of Hoover Dam, the angle of elevation to the top of the dam is 33 degrees. Find the height of the dam to the nearest meter.

Trigonometry Word Problems Students solve word problems using sine, cosine, and tangent. The terms angle of elevation and angle of depression are also introduced in this lesson. Example: Neil sees a rocket at an angle of elevation of 11°. If Neil is located 5 miles from the rocket’s launchpad, how high is the rocket? Round your answer to the nearest hundredth.

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Word Problems- Angle of Depression and Elevation

Andymath.com features free videos, notes, and practice problems with answers! Printable pages make math easy. Are you ready to be a mathmagician?

\(\textbf{1)}\) You are standing 10 feet from the base of a tree. You look up at the top of the tree with an angle of elevation of 60 degrees. How tall is the tree? Show Answer The answer is \(\approx 17.3 \) feet Show Work \(\,\,\,\,\,\,\tan{60}=\frac{x}{10}\) \(\,\,\,\,\,\,\sqrt{3}=\frac{x}{10}\) \(\,\,\,\,\,\,10\sqrt{3}=x\) The answer is \(\approx 17.3 \) feet

\(\textbf{2)}\) you are in a hot air balloon. you look at steve with an angle of depression of 30 degrees. your elevation is 1200 feet. how far apart are you and steve show answer the answer is \( 2400 \) feet show work \(\,\,\,\,\,\,\sin{30}=\frac{1200}{x}\) \(\,\,\,\,\,\,\frac{1}{2}=\frac{1200}{x}\) \(\,\,\,\,\,\,x=2400\) \(\,\,\,\,\,\,\text{the answer is } 2400 \text{ feet}\), \(\textbf{3)}\) you are flying a kite. you let out 40 feet of string at an angle of elevation of 40 degrees. how high up is the kite show answer the answer is \( 25.7 \) feet show work \(\,\,\,\,\,\,\sin{40}=\frac{x}{40}\) \(\,\,\,\,\,\,x=40\sin{40}\) \(\,\,\,\,\,\,x\approx 25.7\) \(\,\,\,\,\,\,\text{the answer is } 25.7 \text{ feet}\), \(\textbf{4)}\) you are on top of a building. you look down on the neighboring building at an angle of depression of 30 degrees. your building is 100 feet tall. the buildings are 30 feet apart. how tall is the other building show answer \(\text{the neighboring building is } 82.7 \text{ feet tall}\) show work \(\,\,\,\,\,\,\tan{30}=\frac{x}{30}\) \(\,\,\,\,\,\,0.5774=\frac{x}{30}\) \(\,\,\,\,\,\,x=30 \cdot 0.5774\) \(\,\,\,\,\,\,x=17.32\) \(\,\,\,\,\,\,h+x=100\) \(\,\,\,\,\,\,h+17.32=100\) \(\text{the neighboring building is } 82.7 \text{ feet tall}\), \(\textbf{5)}\) from the top of a lighthouse, the angle of depression to a boat is 45 degrees. if the lighthouse is 100 feet tall, how far is the boat from the base of the lighthouse show answer the answer is \(100\) feet. show work \(\,\,\,\,\,\,\tan{45}=\frac{x}{100}\) \(\,\,\,\,\,\,1=\frac{x}{100}\) \(\,\,\,\,\,\,x=100\) the boat is \(100\) feet away from the base of the lighthouse., \(\textbf{6)}\) you are flying a kite and hold the string 80 feet long. the angle of elevation to the kite is 50 degrees. how high is the kite above the ground show answer the answer is \(61.3\) feet. show work \(\,\,\,\,\,\,\sin{50}=\frac{x}{80}\) \(\,\,\,\,\,\,x=80\sin{50}\) \(\,\,\,\,\,\,x\approx61.3\) the kite is \(61.3\) feet above the ground., \(\textbf{7)}\) a person looks up at the top of a cliff with an angle of elevation of 20 degrees. if they are standing 200 feet away from the base of the cliff, how tall is the cliff show answer the answer is \(72.8\) feet. show work \(\,\,\,\,\,\,\tan{20}=\frac{x}{200}\) \(\,\,\,\,\,\,x=200\tan{20}\) \(\,\,\,\,\,\,x\approx72.8\) the cliff is \(72.8\) feet tall., \(\textbf{8)}\) from the top of a building, the angle of depression to a car on the street is 40 degrees. if the building is 80 feet tall, how far is the car from the base of the building show answer the answer is \(95.3\) feet. show work \(\,\,\,\,\,\,\tan{40}=\frac{80}{x}\) \(\,\,\,\,\,\,x=\frac{80}{\tan{40}}\) \(\,\,\,\,\,\,x\approx95.3\) the car is \(95.3\) feet away from the base of the building., see related pages\(\), \(\bullet\text{ geometry homepage}\) \(\,\,\,\,\,\,\,\,\text{all the best topics…}\), \(\bullet\text{ right triangle trigonometry}\) \(\,\,\,\,\,\,\,\,\sin{(x)}=\displaystyle\frac{\text{opp}}{\text{hyp}}…\), \(\bullet\text{ angle of depression and elevation}\) \(\,\,\,\,\,\,\,\,\text{angle of depression}=\text{angle of elevation}…\), \(\bullet\text{ convert to radians and to degrees}\) \(\,\,\,\,\,\,\,\,\text{radians} \rightarrow \text{degrees}, \times \displaystyle \frac{180^{\circ}}{\pi}…\), \(\bullet\text{ degrees, minutes and seconds}\) \(\,\,\,\,\,\,\,\,48^{\circ}34’21”…\), \(\bullet\text{ coterminal angles}\) \(\,\,\,\,\,\,\,\,\pm 360^{\circ} \text { or } \pm 2\pi n…\), \(\bullet\text{ reference angles}\) \(\,\,\,\,\,\,\,\,\) \(…\), \(\bullet\text{ find all 6 trig functions}\) \(\,\,\,\,\,\,\,\,\) \(…\), \(\bullet\text{ unit circle}\) \(\,\,\,\,\,\,\,\,\sin{(60^{\circ})}=\displaystyle\frac{\sqrt{3}}{2}…\), \(\bullet\text{ law of sines}\) \(\,\,\,\,\,\,\,\,\displaystyle\frac{\sin{a}}{a}=\frac{\sin{b}}{b}=\frac{\sin{c}}{c}\) \(…\), \(\bullet\text{ area of sas triangles}\) \(\,\,\,\,\,\,\,\,\text{area}=\frac{1}{2}ab \sin{c}\) \(…\), \(\bullet\text{ law of cosines}\) \(\,\,\,\,\,\,\,\,a^2=b^2+c^2-2bc \cos{a}\) \(…\), \(\bullet\text{ area of sss triangles (heron’s formula)}\) \(\,\,\,\,\,\,\,\,\text{area}=\sqrt{s(s-a)(s-b)(s-c)}\) \(…\), \(\bullet\text{ geometric mean}\) \(\,\,\,\,\,\,\,\,x=\sqrt{ab} \text{ or } \displaystyle\frac{a}{x}=\frac{x}{b}…\), \(\bullet\text{ geometric mean- similar right triangles}\) \(\,\,\,\,\,\,\,\,\) \(…\), \(\bullet\text{ inverse trigonmetric functions}\) \(\,\,\,\,\,\,\,\,\sin {\left(cos^{-1}\left(\frac{3}{5}\right)\right)}…\), \(\bullet\text{ sum and difference of angles formulas}\) \(\,\,\,\,\,\,\,\,\sin{(a+b)}=\sin{a}\cos{b}+\cos{a}\sin{b}…\), \(\bullet\text{ double-angle and half-angle formulas}\) \(\,\,\,\,\,\,\,\,\sin{(2a)}=2\sin{(a)}\cos{(a)}…\), \(\bullet\text{ trigonometry-pythagorean identities}\) \(\,\,\,\,\,\,\,\,\sin^2{(x)}+\cos^2{(x)}=1…\), \(\bullet\text{ product-sum identities}\) \(\,\,\,\,\,\,\,\,\cos{\alpha}\cos{\beta}=\left(\displaystyle\frac{\cos{(\alpha+\beta)}+\cos{(\alpha-\beta)}}{2}\right)…\), \(\bullet\text{ cofunction identities}\) \(\,\,\,\,\,\,\,\,\sin{(x)}=\cos{(\frac{\pi}{2}-x)}…\), \(\bullet\text{ proving trigonometric identities}\) \(\,\,\,\,\,\,\,\,\sec{x}-\cos{x}=\displaystyle\frac{\tan^2{x}}{\sec{x}}…\), \(\bullet\text{ graphing trig functions- sin and cos}\) \(\,\,\,\,\,\,\,\,f(x)=a \sin{b(x-c)}+d \) \(…\), \(\bullet\text{ solving trigonometric equations}\) \(\,\,\,\,\,\,\,\,2\cos{(x)}=\sqrt{3}…\), word problems in trigonometry often involve finding the angle of depression or elevation. these angles are formed between a horizontal line of sight and a line of sight downward or upward, respectively. word problems involving angle of depression and elevation are typically covered in high school trigonometry or precalculus classes. we learn about these because it provides a practical application for the concepts learned in this branch of mathematics. these types of problems can be found in fields such as surveying, meteorology, and navigation. common mistakes when solving word problems involving angle of depression and elevation is not understanding which angle is the angle of depression. it is usually not the upper angle in the triangle drawing for the problem. it is the angle off of the horizontal line. usually the angle of depression is equal to the angle of elevation, which is usually the bottom acute angle in the drawing., about andymath.com, andymath.com is a free math website with the mission of helping students, teachers and tutors find helpful notes, useful sample problems with answers including step by step solutions, and other related materials to supplement classroom learning. if you have any requests for additional content, please contact andy at [email protected] . he will promptly add the content. topics cover elementary math , middle school , algebra , geometry , algebra 2/pre-calculus/trig , calculus and probability/statistics . in the future, i hope to add physics and linear algebra content. visit me on youtube , tiktok , instagram and facebook . andymath content has a unique approach to presenting mathematics. the clear explanations, strong visuals mixed with dry humor regularly get millions of views. we are open to collaborations of all types, please contact andy at [email protected] for all enquiries. to offer financial support, visit my patreon page. let’s help students understand the math way of thinking thank you for visiting. how exciting.

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Trigonometry : Solving Word Problems with Trigonometry

Study concepts, example questions & explanations for trigonometry, all trigonometry resources, example questions, example question #1 : solving word problems with trigonometry.

You can draw the following right triangle using the information given by the question:

Since you want to find the height of the platform, you will need to use tangent.

You can draw the following right triangle from the information given by the question.

In order to find the height of the flagpole, you will need to use tangent.

You can draw the following right triangle from the information given in the question:

In order to find out how far up the ladder goes, you will need to use sine.

In right triangle ABC, where angle A measures 90 degrees, side AB measures 15 and side AC measures 36, what is the length of side BC?

This triangle cannot exist.

Example Question #5 : Solving Word Problems With Trigonometry

A support wire is anchored 10 meters up from the base of a flagpole, and the wire makes a 25 o angle with the ground. How long is the wire, w? Round your answer to two decimal places.

23.81 meters

28.31 meters

21.83 meters

To make sense of the problem, start by drawing a diagram. Label the angle of elevation as 25 o , the height between the ground and where the wire hits the flagpole as 10 meters, and our unknown, the length of the wire, as w. 

Now, we just need to solve for w using the information given in the diagram. We need to ask ourselves which parts of a triangle 10 and w are relative to our known angle of 25 o . 10 is opposite this angle, and w is the hypotenuse. Now, ask yourself which trig function(s) relate opposite and hypotenuse. There are two correct options: sine and cosecant. Using sine is probably the most common, but both options are detailed below.

We know that sine of a given angle is equal to the opposite divided by the hypotenuse, and cosecant of an angle is equal to the hypotenuse divided by the opposite (just the reciprocal of the sine function). Therefore:

To solve this problem instead using the cosecant function, we would get:

The reason that we got 23.7 here and 23.81 above is due to differences in rounding in the middle of the problem. 

Example Question #6 : Solving Word Problems With Trigonometry

When the sun is 22 o above the horizon, how long is the shadow cast by a building that is 60 meters high?

To solve this problem, first set up a diagram that shows all of the info given in the problem. 

Next, we need to interpret which side length corresponds to the shadow of the building, which is what the problem is asking us to find. Is it the hypotenuse, or the base of the triangle? Think about when you look at a shadow. When you see a shadow, you are seeing it on something else, like the ground, the sidewalk, or another object. We see the shadow on the ground, which corresponds to the base of our triangle, so that is what we'll be solving for. We'll call this base b.

Therefore the shadow cast by the building is 150 meters long.

If you got one of the incorrect answers, you may have used sine or cosine instead of tangent, or you may have used the tangent function but inverted the fraction (adjacent over opposite instead of opposite over adjacent.)

Example Question #7 : Solving Word Problems With Trigonometry

From the top of a lighthouse that sits 105 meters above the sea, the angle of depression of a boat is 19 o . How far from the boat is the top of the lighthouse?

423.18 meters

318.18 meters

36.15 meters

110.53 meters

To solve this problem, we need to create a diagram, but in order to create that diagram, we need to understand the vocabulary that is being used in this question. The following diagram clarifies the difference between an angle of depression (an angle that looks downward; relevant to our problem) and the angle of elevation (an angle that looks upward; relevant to other problems, but not this specific one.) Imagine that the top of the blue altitude line is the top of the lighthouse, the green line labelled GroundHorizon is sea level, and point B is where the boat is.

Merging together the given info and this diagram, we know that the angle of depression is 19 o  and and the altitude (blue line) is 105 meters. While the blue line is drawn on the left hand side in the diagram, we can assume is it is the same as the right hand side. Next, we need to think of the trig function that relates the given angle, the given side, and the side we want to solve for. The altitude or blue line is opposite the known angle, and we want to find the distance between the boat (point B) and the top of the lighthouse. That means that we want to determine the length of the hypotenuse, or red line labelled SlantRange. The sine function relates opposite and hypotenuse, so we'll use that here. We get:

Example Question #8 : Solving Word Problems With Trigonometry

Angelina just got a new car, and she wants to ride it to the top of a mountain and visit a lookout point. If she drives 4000 meters along a road that is inclined 22 o to the horizontal, how high above her starting point is she when she arrives at the lookout?

9.37 meters

1480 meters

3708.74 meters

10677.87 meters

1616.1 meters

As with other trig problems, begin with a sketch of a diagram of the given and sought after information.

Angelina and her car start at the bottom left of the diagram. The road she is driving on is the hypotenuse of our triangle, and the angle of the road relative to flat ground is 22 o . Because we want to find the change in height (also called elevation), we want to determine the difference between her ending and starting heights, which is labelled x in the diagram. Next, consider which trig function relates together an angle and the sides opposite and hypotenuse relative to it; the correct one is sine. Then, set up:

Therefore the change in height between Angelina's starting and ending points is 1480 meters. 

Example Question #9 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 50 feet apart. The shorter building is 40 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 48 o . How high is the taller building?

To solve this problem, let's start by drawing a diagram of the two buildings, the distance in between them, and the angle between the tops of the two buildings. Then, label in the given lengths and angle. 

Example Question #10 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 80 feet apart. The shorter building is 55 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 32 o . How high is the taller building?

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Lesson Plan: Angles of Elevation and Depression Mathematics • Second Year of Secondary School

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This lesson plan includes the objectives, prerequisites, and exclusions of the lesson teaching students how to solve real-world problems that involve angles of elevation and depression.

Students will be able to

• recognize which angle is an angle of depression and which angle is an angle of elevation,
• sketch a diagram from a real-world problem involving an angle of elevation or depression and use this to help solve the problem,
• use right trigonometry along with the law of sines and cosines to solve problems involving one or more angles of elevation.

Prerequisites

Students should already be familiar with

• how to describe directions using a combination of angles and north, south, east, and west,
• using right triangle trigonometry to find unknown angle measures and side lengths,
• the law of sines and the law of cosines and how to use them to calculate unknown angle measures and side lengths.

Students will not cover

• problems that do not involve angles of elevation or depression.

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Trigonometry - Angles of Elevation and Depression

Subject: Mathematics

Age range: 14-16

Resource type: Lesson (complete)

Last updated

1 September 2024

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This resource includes:

• A PowerPoint that can be used with a class,
• A scaffolded student version of the PowerPoint with gaps for the students to fill in,
• A differentiated worksheet with answers included.

This resource is suitable for Year 9 and 10 students and is linked to the Australian curriculum. This resource focuses on understanding what angles of elevation and depression are, drawing diagrams from written descriptions and answering worded questions involving angles of elevation and depression.

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