Exercise 60 Page 495

Let's use the Polygon Angle-Sum Theorem, which tells us that the measures of the interior angles of an n-sided polygon must add up to (n-2)180^(∘). Since a quadrilateral is a 4-sided polygon, the measures must add up to (4-2)180^(∘)=360^(∘). This gives us the following equation. 102^(∘) + x + 46^(∘) + 130^(∘) = 360^(∘) Let's solve the equation for x.

102^(∘) + x + 46^(∘) + 130^(∘) = 360^(∘)

AddTerms

Add terms

278^(∘) + x = 360^(∘)

SubEqn

LHS-278^(∘)=RHS-278^(∘)

x = 82^(∘)

Let's use the Polygon Angle-Sum Theorem again. Since our polygon has 6 sides, the measures must add up to (6-2)180^(∘)=720^(∘). Applying the theorem gives us the following equation. (6x+10^(∘))+(9x+4^(∘))+(7x+2^(∘)) +(12x+8^(∘))+(8x-16^(∘))+10x ⇓ =720^(∘) Let's solve the equation for x.

(6x+10^(∘))+(9x+4^(∘))+(7x+2^(∘)) +(12x-8^(∘))+(8x-16^(∘))+10x=720^(∘)

AddTerms

Add terms

52x -8^(∘) = 720^(∘)

AddEqn

LHS+8^(∘)=RHS+8^(∘)

52x = 728^(∘)

DivEqn

.LHS /52.=.RHS /52.

x = 14^(∘)

Our first step will be to express the unknown angle of the pentagon in terms of x. Notice that this angle and the angle with a measure of x are alternate exterior angles. Since the segments cut by the transversal are parallel, by the Alternate Exterior Angles Theorem, they are congruent.

Now let's use the Polygon Angle-Sum Theorem to find the value of x. Since our polygon has 5 sides, the measures must add up to (5-2)180^(∘)=540^(∘). Applying the theorem gives us the following equation. x + 113^(∘)+97^(∘)+90^(∘)+123^(∘) = 540^(∘) Let's solve the equation for x.

x + 113^(∘)+97^(∘)+90^(∘)+123^(∘) = 540^(∘)

AddTerms

Add terms

x + 423^(∘) = 540^(∘)

SubEqn

LHS-423^(∘)=RHS-423^(∘)

x = 117^(∘)