Exercise 93 Page 636

a Let's copy the diagram and add the given information. The fact that AD is tangent to ⊙ C means ∠ ADB is a right angle.

To find the area of ⊙ C we have to determine the circle's radius. Notice that △ ABD is a right triangle where the hypotenuse and leg are known. This means we can determine the length of BD with the Pythagorean Theorem.

a^2+b^2=c^2

SubstituteII

a= 9, c= 15

9^2+b^2= 15^2

▼

Solve for b

CalcPow

Calculate power

81+b^2=225

SubEqn

LHS-81=RHS-81

b^2=144

SqrtEqn

sqrt(LHS)=sqrt(RHS)

b=± 12

b > 0

b=12

As we can see, BD is 12 cm. Since this is the circle's diameter, the circle's radius must be r= 6 cm. We can determine the area of the circle. Area: π( 6)^2 ≈ 113.1 cm^2

b Let's add the new information to the diagram. We will also mark EB.

Since BD is the circle's diameter, DEB must be a semicircle. Since a semicircle makes up 180^(∘), the measure of EB must be the difference between 180^(∘) and mEB.

mEB: 180^(∘)- 30^(∘) = 150^(∘) To find the measure of AD we recognize the fact that if the radius of ⊙ C is 10 cm, then its diameter must be 20 cm. Additionally, ED is the intercepted arc to the inscribed angle ∠ EBD. This must mean that ∠ EBD is half the measure of ED. m∠ EBD: 30^(∘)/2=15^(∘) Let's add this information to the diagram.

Now we have enough information to calculate AD using the tangent ratio.

tan θ = Opposite/Adjacent

SubstituteValues

Substitute values

tan 15^(∘) = AD/20

▼

Solve for AD

MultEqn

LHS * 20=RHS* 20

20tan 15^(∘) = AD

RearrangeEqn

Rearrange equation

AD=20tan 15^(∘)

UseCalc

Use a calculator

AD=5.35898...

RoundDec

Round to 2 decimal place(s)

AD≈ 5.36

c Let's add the given information to the diagram. We will also mark EB.

Since BC is the radius of ⊙ C, its diameter must be 14 cm. Also, examining the diagram, we see that EB is the intercepted arc to ∠ BDE. With this information we can find ∠ BDE.

m∠ BDE: 86^(∘)/2=43^(∘) Additionally, BD is the intercepted arc to the inscribed angle ∠ BED. Since BD is a semicircle, it must be that m∠ BED=90^(∘). Let's add this information to the diagram.

Notice that EB is the opposite leg to the 43^(∘) angle in a right triangle. Since we know the hypotenuse of this triangle, we can find EB with the sine ratio.

sin θ = Opposite/Hypotenuse