Exercise 61 Page 679

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.

Now let's look at the given picture. We will name the missing side lengths and angle measure.

First, we can evaluate the value of x by using the Triangle Angle Sum Theorem.

51^(∘)+ 43^(∘)+ x=180^(∘)

AddTerms

Add terms

94^(∘)+x=180^(∘)

SubEqn

LHS-94^(∘)=RHS-94^(∘)

x=86^(∘)

The third angle in this triangle has a measure of 86^(∘).

To find the other two side lengths, we will use the Law of Sines. Let's write a proportion. sin 43^(∘)/y=sin 51^(∘)/103=sin 86^(∘)/z Let's start with evaluating the value of y using cross multiplication.

sin 43^(∘)/y=sin 51^(∘)/103

CrossMult

Cross multiply

103sin43^(∘)=ysin51^(∘)

DivEqn

.LHS /sin51^(∘).=.RHS /sin51^(∘).

103sin43^(∘)/sin51^(∘)=y

RearrangeEqn

Rearrange equation

y=103sin43^(∘)/sin51^(∘)

UseCalc

Use a calculator

y=90.3894...

RoundDec

Round to 1 decimal place(s)

y≈ 90.4

Now we will solve for z.

sin 51^(∘)/103=sin 86^(∘)/z

CrossMult

Cross multiply

zsin51^(∘)=103sin86^(∘)

DivEqn

.LHS /sin51^(∘).=.RHS /sin51^(∘).

z=103sin86^(∘)/sin51^(∘)

UseCalc

Use a calculator

z=132.2133...

RoundDec

Round to 1 decimal place(s)

z≈ 132.2

Let's add this information to our diagram.

Now we can evaluate the perimeter of the triangle by adding all the side lengths. Remember that this will be an approximation as we are using approximate side lengths. 103+ 90.4+ 132.2=325.6 The perimeter is approximately 325.6 units.