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.
