Exercise 45 Page 388

Practice makes perfect

a The two angles are complementary because they form a right angle when added. Therefore, we can set the sum of their expressions equal to 90^(∘).

(4x-2^(∘))+x=90^(∘) Let's solve this equation for x

(4x-2^(∘))+x=90^(∘)

▼

Solve for x

RemovePar

Remove parentheses

4x-2^(∘)+x=90^(∘)

AddTerms

Add terms

5x-2^(∘)=90^(∘)

AddEqn

LHS+2^(∘)=RHS+2^(∘)

5x=92^(∘)

DivEqn

.LHS /5.=.RHS /5.

x=18.4^(∘)

b By the Triangle Angle Sum Theorem, we know that the sum of a triangles angles equals 180^(∘).

(2m+3^(∘)) + (m+9^(∘)) + (m-1^(∘))=180^(∘) Let's solve for m in this equation.

(2m+3^(∘)) + (m+9^(∘)) + (m-1^(∘))=180^(∘)

▼

Solve for m

RemovePar

Remove parentheses

2m+3^(∘) + m+9^(∘) + m-1^(∘)=180^(∘)

AddSubTerms

Add and subtract terms

4m+11^(∘)=180^(∘)

SubEqn

LHS-11^(∘)=RHS-11^(∘)

4m=169^(∘)

DivEqn

.LHS /4.=.RHS /4.

m=42.25^(∘)

c The two angles are vertical angles which means they are congruent by the Vertical Angles Theorem.

7k-6^(∘) = 3k+18^(∘) Let's solve for the unknown angle.

7k-6^(∘) = 3k+18^(∘)

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Solve for k

AddEqn

LHS+6^(∘)=RHS+6^(∘)

7k = 3k+24^(∘)

SubEqn

LHS-3k=RHS-3k

4k = 24^(∘)

DivEqn

.LHS /4.=.RHS /4.

k = 6^(∘)

d In the diagram, we can make out two triangles. To calculate x, we first have to show that they are similar. This means they have at least two pairs of congruent angles. First, we notice that they share an angle

We can also identify a pair of corresponding angles and since the two sides cut by the third side are parallel, we know by the Corresponding Angles Theorem that they are congruent.

Since two pairs of angles are congruent, we can claim that the triangles are similar by the AA Similarity condition. Having proved the triangles similar, we can write and solve an equation for x.

x/16=8/13

▼

Solve for x

MultEqn

LHS * 16=RHS* 16

x=8/13* 16

MoveRightFacToNum

a/c* b = a* b/c

x=128/13

UseCalc

Use a calculator

x=9.84615...