Exercise 32 Page 341

There are two parts to this exercise. First, we need to find the value of x. Then we can use that value to solve for the measures of the angles.

Finding x

Let's use the information given to set up and solve an equation for x. On the diagram, we are given the measure of an interior angle of a triangle. We are also given expressions for another interior angle and an exterior angle of the same triangle.

The Exterior Angle Theorem gives a relationship between these angles.

Exterior Angle Theorem

The measure of the exterior angle is the sum of the measures of the two remote interior angles.

This relationship makes it possible to set up and solve an equation for x.

37+(3x+47)=5x+62

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Solve for x

RemovePar

Remove parentheses

37+3x+47=5x+62

AddTerms

Add terms

3x+84=5x+62

SubEqn

LHS-62=RHS-62

3x+22=5x

SubEqn

LHS-3x=RHS-3x

22=2x

DivEqn

.LHS /2.=.RHS /2.

11=x

RearrangeEqn

Rearrange equation

x=11

The value of x on the diagram is 11.

Finding the Angle Measures

We can substitute the value we found for x, x= 11, into the expressions to find the measures of the angles. 3x+47 ⇒ & 3( 11)+47=80^(∘) 5x+62 ⇒ & 5( 11)+62=117^(∘)