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 48 Page 677

Find the missing sides using the Law of Sines.

≈ 14.7

Practice makes perfect

Let's begin with recalling the Law of Sines. If △ ABC has lengths of a, b, and c and angle measures of A, B, and C, then we can write that the ratios of the sine of an angle to the side opposite this angle are equal.

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, sides, and angles.

Now let's evaluate the missing angle measures. To do this we will use the Triangle Angle Sum Theorem.

We will start with evaluating the value of x. x+ 98^(∘)+ 27^(∘)=180^(∘) ⇓ x=55^(∘) Next, we will find the value of y.

y+40^(∘)+68^(∘)=180^(∘) ⇓ y=72^(∘) Let's add this information to our picture.

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

To find these values, we will create a proportion using the Law of Sines. sin 55^(∘)/a=sin 27^(∘)/2.3=sin 98^(∘)/d Let's start with evaluating the value of a using cross multiplication.
sin 55^(∘)/a=sin 27^(∘)/2.3
2.3sin55^(∘)=asin27^(∘)
2.3sin55^(∘)/sin27^(∘)=a
a=2.3sin55^(∘)/sin27^(∘)
a=4.1499...
a≈ 4.1
Now we will solve for d.
sin 27^(∘)/2.3=sin 98^(∘)/d
dsin27^(∘)=2.3sin98^(∘)
d=2.3sin98^(∘)/sin27^(∘)
d=5.0168...
d≈ 5
Let's add this information to our picture.

The next step will be to evaluate the values of b and c.

To do this, we will create a proportion using the Law of Sines. sin 40^(∘)/b=sin72^(∘)/5=sin68^(∘)/c First we will evaluate the value of b using cross multiplication.
sin 40^(∘)/b=sin72^(∘)/5
5sin40^(∘)=bsin72^(∘)
5sin40^(∘)/sin72^(∘)=b
b=5sin40^(∘)/sin72^(∘)
b=3.3793...
b≈ 3.4
This time we will solve for c.
sin72^(∘)/5=sin68^(∘)/c
csin72^(∘)=5sin68^(∘)
c=5sin68^(∘)/sin72^(∘)
c=4.8744...
c≈ 4.9
Let's add this information to our picture.

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

Remember that this will be an approximation as we are using approximate side lengths. 4.1+ 2.3+3.4+4.9=14.7 The perimeter of this figure is approximately 14.7 units.