McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. The Law of Sines and Law of Cosines
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Exercise 49 Page 677

Find the missing sides using the Law of Cosines and the Law of Sines.

≈ 275.1

Practice makes perfect

In our exercise we are given a quadrilateral and asked to evaluate its perimeter. Let's take a look at the given picture. We will label the missing diagonal and sides.

First, we can evaluate the value of d. To do this, we will use the Law of Cosines.

Let's write an equation using this law. d^2= 63^2+ 60^2-2( 63)( 60)cos 48^(∘) Now we will solve the equation. Notice that since d represents a length, we will consider only the positive case when taking a square root of d^2.
d^2=63^2+60^2-2(63)(60)cos 48^(∘)
Solve for d
d^2=3969+3600-2(63)(60)cos 48^(∘)
d^2=3969+3600-7560cos 48^(∘)
d^2=7569-7560cos 48^(∘)
sqrt(d^2)=sqrt(7569-7560cos 48^(∘))
d=sqrt(7569-7560cos 48^(∘))
d=50.1036...
d≈ 50.1
Let's add this information to our picture.

Next, we can find the values of a and b.

First, we will evaluate the measure of the angle that lies opposite a, which we can call A. To do this, we will use the Triangle Angle Sum Theorem. A+37^(∘)+87^(∘)=180^(∘) ⇓ A=56^(∘) Let's add this measure to our picture.

To find the missing values, we will create a proportion using the Law of Sines. sin 56^(∘)/a=sin37^(∘)/50.1=sin87^(∘)/b Let's start with evaluating the value of a using cross multiplication.
sin 56^(∘)/a=sin37^(∘)/50.1
50.1sin56^(∘)=asin37^(∘)
50.1sin56^(∘)/sin37^(∘)=a
a=50.1sin56^(∘)/sin37^(∘)
a=69.0158...
a≈ 69
Now we will solve for b.
sin37^(∘)/50.1=sin87^(∘)/b
bsin37^(∘)=50.1sin87^(∘)
b=50.1sin87^(∘)/sin37^(∘)
b=83.1340...
b≈ 83.1
Let's add this information to our picture.

Now we can evaluate the perimeter of the figure by adding all the side lengths.

Remember that this will be an approximation as we are using approximate side lengths. 63+ 60+69+83.1=275.1 The perimeter of this figure is approximately 275.1 units.