Lesson Video: Angles of Elevation and Depression | Nagwa Lesson Video: Angles of Elevation and Depression | Nagwa

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

In this video, we will learn how to solve real-world problems that involve angles of elevation and depression.

19:14

### Video Transcript

In this video, we will learn how to solve real-world problems that involve angles of elevation and depression. Before we begin, we should already be familiar with using trigonometry in right triangles to find unknown angles and sides and applying the laws of sines and cosines in non-right triangles.

Let’s start by discussing what these terms angle of elevation and angle of depression mean. Suppose you’re standing on the edge of a cliff. An angle of elevation is the angle formed between the horizontal and your eye line as you look up towards something. For example, if you’re looking up towards the sun, then there’s an angle of elevation formed between the horizon and your line of sight here. An angle of depression on the other hand is the angle formed between the horizontal and your eye line as you look down to something. So, for example, if you’re looking down towards a boat out at sea, there’s an angle of depression here. The key thing to remember is that in each case the angle is formed between the horizontal and the line of sight.

Now, problems involving angles of elevation and depression can often be answered using right triangles. If we sketch in a vertical line, then we have right triangles formed by the vertical, the horizontal, and the line of sight. That’s these triangles here. And the angle of elevation or the angle of depression is one of the angles in this triangle. So the skills of applying trigonometry in right triangles are really useful here. In more complex problems, we may need to apply the law of sines or the law of cosines to non-right triangles as we’ll see in our examples.

Each of the problems we consider in this video will be practical problems. And in some cases, we’ll need to draw a diagram ourselves from a worded description. This is a really important skill. And as always, we need to make sure we read all the information in the question carefully. Our first problem, though, has a diagram already drawn for us.

The distance between two buildings is 40 meters. The top of building 𝐶𝐷 has an angle of elevation of 30 degrees, measured from the top of building 𝐴𝐵. If the height of building 𝐴𝐵 equals 30 meters and the bases of the two buildings are on the same horizontal plane, then the height of 𝐶𝐷 to the nearest meter equals how many meters.

So we’ve been given the diagram to go alongside this problem. And all the information in the question has been labeled on it. We have an angle of elevation here of 30 degrees formed between the horizontal and the line of sight as we look up from building 𝐴𝐵 to building 𝐶𝐷. We also know the height of building 𝐴𝐵, it’s 30 meters, and the horizontal distance between the two buildings, 40 meters. We’re told that the two buildings are on the same horizontal plane, which simply means we can assume that the ground between them is flat.

What we’re looking to calculate is the height of building 𝐶𝐷. From the diagram, we can see that this will be composed of two lengths, a portion which is the same height as building 𝐴𝐵, so that’s 30 meters, and a portion which is this currently unknown length here, which we can think of as 𝑥 meters. We can also see that this length 𝑥 is one side in a right triangle. And in this triangle, we know one other side of 40 meters and one angle of 30 degrees. We can therefore apply right-angle trigonometry to calculate 𝑥.

Labeling the sides of this triangle in relation to the 30-degree angle, we can see that we know the adjacent and we want to calculate the opposite. So it’s the tan ratio that we’re going to use. Recalling that tan is opposite over adjacent, we have that tan of 30 degrees is equal to 𝑥 over 40. Multiplying both sides of this equation by 40, we have that 𝑥 is equal to 40 multiplied by tan of 30 degrees. And evaluating this on a calculator, ensuring our calculator is in degree mode, we find that 𝑥 is equal to 23.0940 continuing.

We haven’t quite finished though because we need the total height of building 𝐶𝐷. So we need to add on the additional length of 30 meters. Doing so gives 53.0940. We’re asked to give our answer to the nearest meter. So as the digit in the first decimal place is a zero, we round to 53. So by applying trigonometry in the right triangle formed by the horizontal, the vertical, and the line of sight, we found that the height of building 𝐶𝐷 to the nearest meter is 53 meter.

In our next example, we’ll need to draw a diagram ourselves from a worded description of a problem involving angles of elevation and depression.

A building is eight meters tall. The angle of elevation from the top of the building to the top of the tree is 44 degrees and the angle of depression from the top of the building to the base of the tree is 58 degrees. Find the distance between the base of the building and the base of the tree giving the answer to two decimal places.

So let’s begin by sketching this problem. First, we have a building which is eight meters tall. We’re then told about an angle of elevation from the top of this building to the top of a tree. Remember, an angle of elevation is measured from the horizontal up towards something. So this tree is taller than the building. So we add the tree and the angle of elevation of 44 degrees. Next, we’re told that the angle of depression from the top of the building to the base of the tree is 58 degrees. We need to be careful when labeling this angle. Remember that an angle of depression is measured from the horizontal down towards something, so the angle of 58 degrees is this angle here.

What we’re looking to calculate is the distance between the base of the building and the base of the tree, which is this length here. Now, looking carefully at our diagram, we can see that we have a right triangle. And the length we’re looking to calculate, which we’ll call 𝑦 meters, is one of the sides. We also know another of the sides. It’s the height of the building, eight meters. We can also work out one of the angles inside our triangle. If we subtract the angle of depression of 58 degrees from a right angle of 90 degrees, we can find the top angle in our triangle. It’s 32 degrees. So we now have a right triangle in which we know one length and one other angle, and we wish to calculate a second length.

Labeling the sides of the triangle in relation to the 32-degree angle, we know the adjacent and we want to calculate the opposite. So we’re going to use the tan ratio. Recall, the tan ratio is opposite over adjacent, so we have tan of 32 degrees is equal to 𝑦 over eight. Multiplying both sides of this equation by eight, we have that 𝑦 is equal to eight tan 32 degrees. Evaluating this on a calculator, which must be in degree mode, we have 4.9989.

We’re asked to give our answer to two decimal places. So starting with the eight in the third decimal place and rounding up, we have 5.00 meters. So we’ve completed the problem. The distance between the base of the building and the base of the tree to two decimal places is 5.00 meters.

Now, in fact, in this problem, we didn’t actually need to use the first piece of information we were given, the angle of elevation. The trigonometry involved was quite straightforward. What this problem was really testing was our understanding of angles of elevation and depression and whether we can accurately draw a diagram based on a worded description.

In our next example, we’ll apply our knowledge of angles of elevation and depression to a more complex problem.

A boat was sailing in a straight line toward a rock with a uniform velocity of 96 meters per minute. At one point, the angle of elevation to the top of the rock was 39 degrees. And three minutes later, it became 44 degrees. Find the height of the rock giving the answer to the nearest meter.

So, firstly, we need to think carefully about how we can draw a diagram to represent this problem. We have a rock and a boat, which we’ll simplify as a pink dot. Initially, we’re told that the angle of elevation to the top of the rock is 39 degrees. Remember, an angle of elevation is measured from the horizontal to the line of sight when we look up towards something. So that’s this angle here. The boat then continues on its journey in a straight line towards the rock. And three minutes later, the new angle of elevation is 44 degrees. So from the new position of the boat, that’s this angle here.

We can see that we have a triangle formed by the two lines of sight from the two positions to the top of the rock and the distance traveled by the boat. We can work this distance out because we know the boat sails with a uniform velocity of 96 meters per minute and the journey takes three minutes. If the boat covers 96 meters in each minute, it will cover 96 times three, that’s 288 meters, in three minutes.

Now, we’re looking to find the height of the rock. That’s this length here. And it isn’t a side in our triangle. However, it is a side length in the right triangle formed by the horizontal, the vertical, and the second line of sight. And we know one angle of 44 degrees in this triangle. This side formed by the line of sight is shared with the first triangle. So our approach is going to be to use the first triangle to try and work out this shared side and then use this shared side and the angle of 44 degrees in the right triangle to calculate the height of the rock.

Let’s look at this non-right triangle more closely then. We have an angle of 39 degrees, and we can work out each of the other two angles. For example, the obtuse angle is on a straight line with the angle of elevation of 44 degrees. So we can calculate it by subtracting 44 from 180, which gives 136 degrees. To calculate the final angle, we can use the angle sum in a triangle. 180 minus the other two angles of 136 and 39 gives five. So we have all three angles in this triangle.

If we want to work out a side length in a non-right triangle in which we know all three angles and one other side length, we can apply the law of sines, which tells us that the ratio between a side and the sine of its opposite angle is constant throughout the triangle. 𝑎 over sin 𝐴 equals 𝑏 over sin 𝐵, which is equal to 𝑐 over sin 𝐶, where the lowercase letters represent sides and the uppercase letters represent angles.

Our side, the side highlighted in pink, which we can call 𝑥 meters, is opposite the angle of 39 degrees. And then we also know a side of 288 meters, which is opposite the angle of five degrees. So applying the law of sines, we have 𝑥 over sin of 39 degrees equals 288 over sin of five degrees. To solve this equation for 𝑥, we can multiply both sides by sin of 39 degrees, giving 𝑥 equals 288 sin 39 degrees over sin five degrees. Using a calculator, that evaluates to 2079.544. But we’ll keep the value as accurate as possible for now.

Now that we know the length of the shared side, we can consider the right triangle which contains the height we’re looking to calculate, which we can now call 𝑦 meters. Labeling the sides of this triangle in relation to the angle of 44 degrees, we want to calculate the opposite. And we know the hypotenuse. So recalling SOH CAH TOA, it is the sine ratio that we want to use. Remember, sine is equal to opposite over hypotenuse. So substituting, we have sin of 44 degrees is equal to the opposite 𝑦 over the hypotenuse 𝑥, which we’ve calculated to be 2079.544.

We can then multiply through by this value of 2079.544 to give 𝑦 equals 2079.544 multiplied by sin of 44 degrees. Now, of course, it makes sense to have kept that value of 2079.544 and so on our calculator display and then just type “multiplied by sin of 44 degrees” to give an answer that is as accurate as possible. Evaluating this gives 1444.57 continuing. We were asked to give our answer to the nearest meter, so the five in the first decimal place means we’re going to be rounding up, giving a height of 1445 meters to the nearest meter.

In this problem then, we used the law of sines in a non-right triangle and right-angle trigonometry in a right triangle in order to calculate this missing length. A key skill at the start of the question was drawing a diagram to represent the problem.

Let’s now consider one final example, this time involving angles of depression.

The angle of depression of a car parked on the ground from the top of a hill is 48 degrees. A viewpoint is 14 meters vertically below the top of the hill, and the angle of depression to the car is 25 degrees. Find the height of the hill giving the answer to the nearest meter.

So we’ve been given a lot of information but no diagram. We need to begin by drawing a sketch. We have a hill, and then we have a car parked down on the ground some distance away. We’re told the angle of depression from the top of the hill to the car is 48 degrees. Now, this is where we need to be particularly careful. We sketch in the line of sight between the top of the hill and the car and the horizontal. And then, we remember that the angle of depression is measured down from the horizontal to the line of sight. So it’s this angle here, which is 48 degrees.

We also have a viewpoint on the hill, which is 14 meters vertically below the top of the hill. And then the angle of depression from this point to the car is 25 degrees. Again, we need to be careful. This angle is measured from the horizontal down to the line of sight. So it’s this angle here. So we’ve put all the information in the question onto our diagram. What we’re asked to calculate is the height of the hill. That’s this length here, which we can see is composed of two lengths, the length of 14 meters between the top of the hill and the viewpoint and then a second length, which we’ll need to calculate.

Now, the second length is part of a right triangle formed by the horizontal, the vertical, and the line of sight for the second angle of depression. We can work out one of the angles in this triangle. Between the horizontal and vertical, we have a right angle. So this angle here is 19 minus 25 degrees, which is 65 degrees. We want to calculate the side 𝑥, but in order to do this, we need to know a length in our right triangle. Let’s consider instead the green triangle, a non-right triangle but which has a shared side with the triangle we’re interested in.

We know this triangle has one side of length 14 meters. And we can work out some of the other angles. At the top of the triangle, the angle between the horizontal and vertical is 90 degrees, so the internal angle in the triangle is 90 minus 48, which is 42 degrees. Also internal to the triangle, we have an obtuse angle formed by a right angle and the angle of depression of 25 degrees, so this total angle is 115 degrees. The final angle in this triangle can be calculated using the angle sum in a triangle. 180 minus 42 degrees minus 115 degrees is 23 degrees.

If we know one side length and at least two angles in a non-right triangle, we can apply the law of sines to calculate the length of another side. This tells us that the ratio between a side length and the side of its opposite angle is constant. The side length we want to calculate, which we can call 𝑦 meters, is opposite the angle of 42 degrees. And then we know a side length of 14 meters, which is opposite the angle of 23 degrees. So applying the law of sines, we can say that 𝑦 over sin of 42 degrees is equal to 14 over sin of 23 degrees. We can then multiply both sides of this equation by sin of 42 degrees and evaluate to give 23.975. So we’ve found the shared side.

Returning to our right triangle, we now know one other angle of 65 degrees and the hypotenuse of 23.975. So we can use the cosine ratio to calculate 𝑥. Cosine, remember, is adjacent over hypotenuse, so we have cos of 65 degrees equals 𝑥 over 23.975. We can then multiply both sides by 23.975 and evaluate on our calculators to give 𝑥 equals 10.132. Finally, we must remember we’re looking to calculate the total height of the hill. So that’s the sum of this value and the distance of 14 meters. Adding 14 and then rounding to the nearest meter, we find that the height of the hill is 24 meters.

Let’s now summarize the key points from this video. An angle of elevation is the angle formed between the horizontal and the line of sight when we look up towards something, whereas an angle of depression is the angle formed between the horizontal and the line of sight when we look down. Problems involving angles of elevation and depression can be answered using skills relating to triangles, either right-angle trigonometry for right triangles or non–right angle trigonometry such as the law of sines and the law of cosines. We may need to use more than one of these skills within the same problem. If we haven’t been given a diagram as part of the question, then we need to draw our own. And as always, we must make sure we read the information in the question really carefully to make sure our diagram accurately reflects the problem.

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