Exercise 50 Page 677

Let's begin with recalling the Law of Cosines. If △ ABC has lengths of a, b, and c and angle measures of A, B, and C, then we can write equations that relate the side lengths of this triangle and the cosine of one of its angles.

Now let's take a look at the given picture. We are asked to evaluate the perimeter of the given triangle, so we need to know all the side lengths. Let x represent the length of the missing side.

Using the Law of Cosines, we can write an equation for x. x^2= 88^2+ 152^2-2( 88)( 152)cos 39^(∘) Let's solve the equation. Notice that since x represents a length, we will consider only the positive case when taking a square root of x^2.

x^2=88^2+152^2-2(88)(152)cos 39^(∘)

▼

Solve for x

CalcPow

Calculate power

x^2=7744+23104-2(88)(152)cos 39^(∘)

Multiply

Multiply

x^2=7744+23104-26752cos 39^(∘)

AddTerms

Add terms

x^2=30848-26752cos 39^(∘)

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(x^2)=sqrt(30848-26752cos 39^(∘))

SqrtPowToNumber

sqrt(a^2)=a

x=sqrt(30848-26752cos 39^(∘))

UseCalc

Use a calculator

x=100.2885...

RoundDec

Round to 1 decimal place(s)

x≈ 100.3

The length of the missing side is approximately 100.3. Now we can find the perimeter of the triangle by adding all side lengths. Notice that this will be approximation as we are using an approximate side length. P= 100.3+ 88+ 152=340.3 The perimeter of this triangle is approximately 340.3.