Exercise 124 Page 519

Practice makes perfect

a We have been asked to find the value of x in the given diagram.

Notice that the two smaller angles form a linear pair. This means that their measures must add up to 180^(∘). (4x-3^(∘)) + (3x+1^(∘)) = 180^(∘) Let's solve the equation for x.

(4x-3^(∘)) + (3x+1^(∘)) = 180^(∘)

AddTerms

Add terms

7x-2^(∘) = 180^(∘)

AddEqn

LHS+2^(∘)=RHS+2^(∘)

7x = 182^(∘)

DivEqn

.LHS /7.=.RHS /7.

x = 26^(∘)

b Like in Part A, we need to find the value of x in the following diagram.

The two marked angles are a pair of Alternate Exterior Angles. By the Alternate Exterior Angles Theorem, pairs of alternate exterior angles are congruent, so these two angles have the same measure. 5x+6^(∘) = 2x+21^(∘) Let's solve the equation for x.

5x+6^(∘) = 2x+21^(∘)

SubEqn

LHS-2x=RHS-2x

3x+6^(∘) = 21^(∘)

SubEqn

LHS-6^(∘)=RHS-6^(∘)

3x = 15^(∘)

DivEqn

.LHS /3.=.RHS /3.

x = 5^(∘)

c Let's take a look at the next diagram.

The diagram shows a triangle with the measures of all of its interior angles expressed in terms of x. By the Triangle Angle Sum Theorem, these measures will add up to 180^(∘). x+19^(∘) + 4x+28^(∘) + 3x+13^(∘) = 180^(∘) Let's solve the equation for x.

(x+19^(∘))+(4x+28^(∘))+(3x+13^(∘)) = 180^(∘)

AddTerms

Add terms

8x+60^(∘) = 180^(∘)

SubEqn

LHS-60^(∘)=RHS-60^(∘)

8x = 120^(∘)

DivEqn

.LHS /8.=.RHS /8.

x = 15^(∘)

d Now let's take a look at the final diagram.

The Triangle Exterior Angle Theorem tells us that the measure of an exterior angle of a triangle is equal to the sum of its two remote interior angles. This means that for our triangle, the sum of 40^(∘) and 90^(∘) equals 4x-10^(∘). 40^(∘) + 90^(∘) = 4x-10^(∘) Let's solve our equation for x.

40^(∘) + 90^(∘) = 4x-10^(∘)

AddTerms

Add terms

130^(∘)= 4x-10^(∘)

AddEqn

LHS+10^(∘)=RHS+10^(∘)

140^(∘) = 4x

DivEqn

.LHS /4.=.RHS /4.

35^(∘) = x

RearrangeEqn

Rearrange equation