Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
2. Section 12.2
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Exercise 93 Page 764

Practice makes perfect
a Consider the given diagram.

We want to find the length x. In order to do so, we should first find the radius of the C from the diagram.

Radius

We know that m PMQ = 314^(∘). Since this angle, together with ∠ PCQ form a full angle, the measures of these angles have to sum to 360^(∘). m PMQ + m ∠ PCQ = 360^(∘)Substituting the known value of m PMQ allows us to find the measure m ∠ PCQ
m PMQ + m ∠ PCQ = 360^(∘)
314^(∘) + m ∠ PCQ = 360^(∘)
m ∠ PCQ = 46^(∘)
Now let's add the found measure to the diagram and let's focus on △ CRP. It is a right triangle, since PR is tangent to C at P.
Since we want to find the radius of C, we want to find the length CP. Notice that since we know m∠ PCR and length of the side opposite this angle, PR, we can use the tangent ratio to find CP. tan ( m∠ PCR ) = PR/CP Let's substitute 46^(∘) for m ∠ PCR and 5 for PR and solve the equation above for CP.
tan ( m∠ PCR ) = PR/CP
tan ( 46 ^(∘) ) = 5/CP
CP tan ( 46 ^(∘) ) = 5
CP = 5/tan ( 46 ^(∘) )
CP = 4.828443...
CP ≈ 4.83
Let's add the found the measure to the diagram.

Now we can move to finding x.

Finding x

Let's again focus on the △ CRP. Notice that we know both legs of this triangle. Therefore, we can use the Pythagorean Theorem in order to find the hypotenuse. Let's do so!
CP^2 + PR^2 = CR^2
4.83^2 + 5^2 ≈ CR^2
Solve for CR
23.3289 + 25 ≈ CR^2
48.3289≈ CR^2
sqrt(48.3289) ≈ CR
6.951899... ≈ CR
6.95 ≈ CR
CR ≈ 6.95
Finally, notice that QR of length x is part of the larger segment CR. Also, since point Q lies on the circle, its distance from point C is equal to the radius. Therefore, we can use the Segment Addition Postulate to find the length x.

The lengths of the two smaller segments must sum to CR = 6.95. 4.83 + x ≈ 6.95 ⇒ x ≈ 2.12

b We want to find the value of x using the fact that the radius of the following circle is 7 centimeters.
Let's denote the endpoints of the chord from the diagram as A and B and mark the center of the circle and call it C. Next, connect C with A and B. Since the radius of a circle is 7 centimeters, both AC and BC are 7 centimeters long.
Now, let's use the Law of Cosines to relate AB to AC and BC, and the measure of ∠ ACB. AB^2 = AC^2 + BC^2 - 2AC * BC * cos ∠ ACB We know that AC = 7 centimeters and BC = 7 centimeters. Also, since m AB = 102^(∘), we have that the central angle ∠ ACB measures 102^(∘). Let's substitute these values into the above formula to get an equation for AB. We can also substitute x for AB to immediately get an equation for x. Let's do it!
AB^2 = AC^2 + BC^2 - 2AC * BC * cos ∠ ACB
x^2 = 7^2 + 7^2 - 2( 7)( 7)cos 102^(∘)
Simplify right-hand side
x^2 = 49 + 49 -2(7)(7)cos 102^(∘)
x^2 = 98 -2(7)(7)cos 102^(∘)
x^2 = 98 - 98cos 102^(∘)
x = sqrt(98 - 98cos 102^(∘))
x = 10.880043...
x ≈ 10.88
We found that x is about 10.88 centimeters. Note that we only care about the principal root since x, as a length, is positive.