Exercise 11 Page 676

Practice makes perfect

a Let's label the vertices of the segment's endpoints that we see inside the circle.

Notice that BC and DC are both radii of the circle. Therefore these segments are congruent, which means △ BCD must be an isosceles triangle. According to the Base Angles Theorem we know that ∠ B ≅ ∠ D, which means m∠ B= x.

Notice that the central angle ∠ BCD has the same measure as its intercepted arc BD. Let's add that information to the diagram.

Using the Triangle Angle Sum Theorem, we can write an equation containing x. m∠ x+m∠ x+106^(∘)=180^(∘) Let's solve this equation by performing inverse operations.

m∠ x+m∠ x+106^(∘)=180^(∘)

▼

Solve for m∠ x

AddTerms

Add terms

2m∠ x+106^(∘)=180^(∘)

SubEqn

LHS-106^(∘)=RHS-106^(∘)

2m∠ x=74^(∘)

DivEqn

.LHS /2.=.RHS /2.

m∠ x=37^(∘)

b Examining the diagram, we can identify two pairs of consecutive interior angles.

Since the two sides that are cut by transversals are parallel, we know by the Consecutive Interior Angles Theorem that they are supplementary. With this information we can write two equations. x+67^(∘)&=180^(∘) 5y+(3y-16^(∘))&=180^(∘) Let's solve them one at a time.

x+67^(∘)=180^(∘)

SubEqn

LHS-67^(∘)=RHS-67^(∘)

x=113^(∘)

Let's solve the second equation as well.

5y+(3y-16^(∘))=180^(∘)

▼

Solve for y

RemovePar

Remove parentheses

5y+3y-16^(∘)=180^(∘)

AddTerms

Add terms

8y-16^(∘)=180^(∘)

AddEqn

LHS+16^(∘)=RHS+16^(∘)

8y=196^(∘)

DivEqn

.LHS /8.=.RHS /8.

y=24.5^(∘)

c With the given information we can use the Law of Sines to solve for x.

a/sin A= b/sin B

SubstituteValues

Substitute values

x/sin 57^(∘)= 9/sin 73^(∘)

▼

Solve for x

MultEqn

LHS * sin 57^(∘)=RHS* sin 57^(∘)

x= 9/sin 73^(∘) * sin 57^(∘)

MoveRightFacToNum

a/c* b = a* b/c

x= 9sin 57^(∘)/sin 73^(∘)

UseCalc

Use a calculator

x=7.89291...

RoundDec

Round to 1 decimal place(s)

x≈ 7.9

To find the length of y we must know the side's opposite angle. This will allow us to use the Law of Sines again. Since we know two angles in the triangle, we can use the Triangle Angle Sum Theorem to determine the remaining angle.

m∠ Y +73^(∘)+57^(∘) = 180^(∘)

AddTerms

Add terms

m∠ Y +130^(∘) = 180^(∘)

SubEqn

LHS-130^(∘)=RHS-130^(∘)

m∠ Y = 50^(∘)

Let's add this angle to the diagram.