### Video Transcript

In this video, we will learn how to
use the cosine and sine law to solve real-world problems. As we’re focusing on the
applications of these laws, we’ll assume at this point familiarity with the laws
themselves. And we won’t explain the basics of
these laws or how to prove them in this video.

Firstly though, a quick reminder of
what the sine and cosine laws actually are and when they can be used. Suppose we have a triangle 𝐴𝐵𝐶
labeled as follows. Remember that the convention is to
use uppercase letters to represent angles and to use lowercase letters to represent
sides, with sides always being opposite the angle with the same letter. None of the angles in this triangle
are necessarily right angles. So the first key point we need to
recall is that the sine and cosine laws allow us to calculate side lengths and angle
measures in nonright triangles.

The law of sines or the sine law
first of all then. This tells us that, in any
triangle, the ratio between each side length and the sine of its opposite angle is
constant, which we can write as 𝑎 over sin 𝐴 equals 𝑏 over sin 𝐵, which is equal
to 𝑐 over sin 𝐶. And remember, the lowercase letters
represent side lengths and the uppercase letters represent angles.

We don’t need to use all three
parts of the sine law, just two parts of this equality. The information needed in order to
apply the law of sines is two angles and their opposite sides. So we can recognize the need for
the law of sines when we’re working with opposite pairs of information.

This first version of the law of
sines is particularly useful when calculating the length of a missing side, as the
sides are in the numerators of the fractions, so less rearrangement is required. We do also have a reciprocal
version, which is particularly helpful for calculating the measure of an angle. sin
𝐴 over 𝑎 equals sin 𝐵 over 𝑏, which is equal to sin 𝐶 over 𝑐. So that’s the law of sines, and now
let’s consider the law of cosines or the cosine law.

The most common form in which you
see this written is as follows. 𝑎 squared equals 𝑏 squared plus
𝑐 squared minus two 𝑏𝑐 cos of 𝐴. We use it to calculate a side
length, in this case the length of side 𝑎, when we know the other two sides of the
triangle and the included angle. So we know the sides 𝑏 and 𝑐 and
the angle 𝐴.

We can also rearrange the cosine
law to a form which enables us to calculate any angle in the triangle if we know all
three side lengths. The rearrangement is
straightforward, and it gives cos of 𝐴 is equal to 𝑏 squared plus 𝑐 squared minus
𝑎 squared all over two 𝑏𝑐. In this case, as we said, we know
all three side lengths and we wish to calculate one of the angles.

Again, the proofs and the basic
application of these laws are not going to be considered in this video. Instead, we’re going to focus on
applying these laws to some real-life problems. Let’s look at our first
question.

A plane travels 800 meters along
the runway before taking off at an angle of 10 degrees. It travels a further 1,000 meters
at this angle as seen in the figure. Work out the distance of the plane
from its starting point. Give your answer to two decimal
places.

Looking at the diagram, we can see
that we have a triangle. We want to calculate the distance
of the plane from its starting point. That’s this length here, which we
can refer to as 𝑑 meters. We know the lengths of the other
two sides in this triangle. They are 800 meters and 1,000
meters. And using the fact that angles on a
straight line sum to 180 degrees, we can work out the size of this angle here. It’s 180 degrees minus 10 degrees,
which is 170 degrees.

As this is a non-right-angled
triangle, we need to answer this problem using either the law of sines or the law of
cosines. So the first step is to decide
which of these we need. And that will depend on the
specific combination of information we’ve been given and what we want to
calculate.

In this triangle, we know two sides
and the included angle. And we want to calculate the third
side. We recall then that this means we
should be using the law of cosines. Let’s recall the law of
cosines. It’s 𝑎 squared equals 𝑏 squared
plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴. Now, there’s no need to actually
label our triangle using the letters 𝐴, 𝐵, and 𝐶. Instead, we just remember that the
lowercase letters 𝑏 and 𝑐 represent the two sides we know and the capital letter
𝐴 represents the included angle.

So using 800 and 1,000 as the two
side lengths 𝑏 and 𝑐 and 170 degrees as the angle 𝐴, we have the equation 𝑑
squared equals 800 squared plus 1,000 squared minus two times 800 times 1,000 times
cos of 170 degrees. We can either type this directly
into our calculator or it may be a good idea to break the calculation down into some
stages. In either case, we arrive at 𝑑
squared equals 3,215,692.405.

Now, we must remember that this is
𝑑 squared. It isn’t 𝑑, so we aren’t
finished. We have to square root in order to
find the value of 𝑑. It’s a really common mistake though
to forget to do this. Square rooting gives 𝑑 equals
1,793.235178. The question asks us to give our
answer to two decimal places. So rounding appropriately, we’ve
worked out the distance of the plane from its starting point. It’s 1,793.24 meters to two decimal
places.

In this example, we were given a
diagram to use. Often this won’t be the case, and
we’ll need to use information from a written description to draw our own diagram to
help us answer the question. Let’s now consider an example of
this.

A ship is sailing due south with a
speed of 36 kilometers per hour. An iceberg lies 24 degrees north of
east. After one hour, the ship is 33
degrees south of west of the iceberg. Find the distance between the ship
and the iceberg at this time, giving the answer to the nearest kilometer.

A lot of the skill involved in this
question is in drawing the diagram. Let’s start with a compass showing
the four directions. We’ll then take each statement
separately and consider how to represent it. Firstly, we know that the ship is
sailing due south. Initially, we’re told that an
iceberg lies 24 degrees north of east of the ship’s starting point.

Now, directly east would be
directly to the right of the ship on our diagram. 24 degrees north of east would mean
the iceberg lies somewhere along the line like this. We’re then told that after one
hour, the ship is 33 degrees south of west of the iceberg. Well, west would be the direction
directly to the left of our iceberg. And using alternate angles in
parallel lines, we know that the angle formed here is 24 degrees.

So the full angle between the
horizontal and the position the ship has now moved to is 33 degrees. And we can now see that we have a
triangle. We can work out the angles in our
triangle. For example, this angle here is the
difference between 33 degrees and 24 degrees. It is nine degrees. We could also work out this angle
here, it’s 24 degrees, plus the angle between south and east, which is 90 degrees,
giving a total of 114 degrees.

The one piece of information we
haven’t used yet is that the ship is traveling at a speed of 36 kilometers per
hour. And we know it takes one hour for
the ship to go from its original position to its new position. The ship will therefore have
traveled 36 kilometers in this time. So we also know one side length in
our triangle.

What we were asked to calculate is
the distance between the ship and the iceberg at this later time. So that’s this side here, which we
can refer to as 𝑑 kilometers. We’ve now set up our diagram, and
we see that we have a non-right-angled triangle, which means we’re going to apply
either the law of sines or the law of cosines. Let’s look at the particular
combination of information we’ve got.

We know an angle of nine degrees
and the opposite side of 36 kilometers. We also know an angle of 114
degrees. And we wish to calculate the
opposite side of 𝑑 kilometers. We therefore have opposite pairs of
sides and angles, which tells us that we should be using the law of sines to answer
this question.

Remember, this tells us that the
ratio between each side length, represented using lowercase letters, and the sine of
its opposite angle, represented using capital letters, is constant. 𝑎 over sin 𝐴 equals 𝑏 over sin
𝐵, which is equal to 𝑐 over sin 𝐶. We only need to use two parts of
this ratio. And there’s no need to label our
triangle using the letters 𝐴, 𝐵, and 𝐶 as long as we’re clear about what they
represent.

Our side 𝑑 is opposite the angle
of 114 degrees, and the side of 36 kilometers is opposite the angle of nine
degrees. So we have 𝑑 over sin of 114
degrees equals 36 over sin of nine degrees. We can solve this equation by
multiplying each side by sin of 114 degrees, which is just a value. And it gives 𝑑 equals 36 sin 114
degrees over sin of nine degrees. Evaluating on a calculator, making
sure our calculator is in degree mode, and we have 210.23267.

The question asks us to give our
answer to the nearest kilometer. So rounding appropriately, we have
that the distance between the ship and the iceberg at this time is 210
kilometers.

We’ve now seen one example each of
using the law of sines and the law of cosines to calculate a side length. In our next example, we’ll see how
we can apply the law of cosines to calculate all the missing angles in a triangle
when we know its three side lengths.

Los Angeles is 1,744 miles from
Chicago, Chicago is 712 miles from New York, and New York is 2,451 miles from Los
Angeles. Find the angles in the triangle
with its vertices as the three cities.

Now, whilst a little bit of
knowledge of the geography of the United States of America might be useful here, it
isn’t essential to answering the problem. We can just draw a triangle using
the three lengths given in the question. And if our triangle turned out to
be upside down, it isn’t the end of the world. The triangle should look a little
something like this, and we can add the three distances.

Now, this triangle certainly
doesn’t look as if it’s a right triangle. So we’re going to need to apply
either the law of sines or the law of cosines to this problem. We know all three of the side
lengths, and we want to calculate each of the angles, which tells us that we should
be using the law of cosines. The rearranged version of this,
which is useful for calculating angles, is cos of 𝐴 equals 𝑏 squared plus 𝑐
squared minus 𝑎 squared all over two 𝑏𝑐. If you can’t remember this, you’ll
have to perform the rearrangement yourself from the law of cosines in its
traditional form.

In this question then, let’s use 𝐴
to represent Los Angeles, 𝐶 to represent Chicago, and 𝐵 to represent New York. We’ll use the corresponding
lowercase letters to represent the opposite sides. To calculate our first angle then,
that’s this angle here, we substitute the relevant values. Giving cos of 𝐴 equals 1,744
squared plus 2,451 squared minus 712 squared all over two multiplied by 1,744
multiplied by 2,451.

We can evaluate this on a
calculator. And then to find the value of 𝐴,
we need to use the inverse cosine function. Doing so gives 𝐴 equals 2.334
degrees. So we’ve found the first angle in
the triangle. And we’ll give our answer to two
decimal places.

To calculate the next angle in this
triangle, which this time we’ll use angle 𝐶, we don’t need to relabel our
triangle. We just need to remember that the
letters 𝑏 and 𝑐 represent the two sides which enclose the angle and the letter 𝑎
represents the opposite side. So we use 1,744 and 712 for the two
sides which enclose the angle and 2,451 for the side which is opposite. This gives cos of 𝐶 equals
negative 0.9901. And again, applying the inverse
cosine function, we find that angle 𝐶 equals 171.939 degrees.

So we found two angles in the
triangle. And in fact, to find the third, we
could subtract the two angles we’ve calculated from 180 degrees. But if we don’t use that method,
that will be a useful check. In exactly the same way then, but
this time using 712 and 2,451 as the two sides which enclose the angle and 1,744 as
the opposite side, we find the measure of angle 𝐵 is 5.726 degrees.

Adding the three angles we’ve
found, now each rounded to two decimal places, does indeed give 180 degrees. So we can have some confidence in
our answer. The measures of the three angles in
the triangle formed by these three cities, each to two decimal places then, are 2.33
degrees, 5.73 degrees, and 171.94 degrees.

In our final example, let’s see how
we can use the law of sines and the law of cosines to solve problems in other
mathematical contexts.

𝑀 is the center of a circle, and
𝐴, 𝐵, and 𝐶 are points on the circumference. If 𝐵𝐶 equals 13 centimeters and
the measure of angle 𝐶𝑀𝐵 is 84 degrees, find the area of the circle 𝑀, giving
the answer to the nearest square centimeter.

We know that the area of a circle
is 𝜋𝑟 squared. So really, this problem is about
finding the radius of this circle. Let’s begin by putting the
information we’ve been given on the diagram. 𝐵𝐶 is 13 centimeters, and the
measure of angle 𝐶𝑀𝐵 is 84 degrees. We don’t know the lengths of 𝑀𝐶
or 𝑀𝐵, but they’re each the radius of the circle.

Now, there are numerous different
approaches we could take. But one approach is to apply the
law of cosines in the triangle 𝐶𝑀𝐵. This states that 𝑎 squared equals
𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴, where 𝑏 and 𝑐 represent two
sides of a triangle and 𝐴 represents the included angle. In our triangle, 𝑎 is 13
centimeters. The angle 𝐴 is 84 degrees. And the two sides which enclose
this angle 𝐴 are each the radius of the circle 𝑟.

We can therefore form an
equation. 13 squared equals 𝑟 squared plus
𝑟 squared minus two 𝑟 squared cos of 84 degrees. We can solve this equation to find
the value of 𝑟 squared, which we’ll then be able to substitute directly into our
area formula. Factorizing the right-hand side of
our equation by 𝑟 squared, we have 169 equals 𝑟 squared multiplied by two minus
two cos of 84 degrees. Dividing through, we have that 𝑟
squared is equal to 169 over two minus two cos 84 degrees. And we’ll keep our value for 𝑟
squared in this exact form.

We can then substitute this value
of 𝑟 squared into the area formula and evaluate on a calculator. Rounding our answer, and we have
that the area of circle 𝑀 to the nearest square centimeter is 296 square
centimeters.

As I mentioned, there are in fact
numerous approaches to this problem, which you can try out yourself if you wish. We could’ve applied the law of
sines in triangle 𝐶𝑀𝐵. Or we could’ve divided it in half
to form two right triangles and then used right-angle trigonometry.

Let’s now summarize the key points
in this video. Firstly, the law of sines and the
law of cosines can be used to calculate side lengths and angle measures in nonright
triangles. The law of sines in either of its
two forms can be used to calculate either a side or an angle when we’re working with
opposite pairs of information. The law of cosines in its first
form can be used to calculate a side length when we know the other two sides and the
included angle. And in its rearranged form, it can
be used to calculate any angle when we know all three sides.

We can apply the law of sines and
the law of cosines to many problems involving triangles. And although we haven’t seen an
example of this, we can also apply both rules within the same problem.