EXAMPLE 2
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EXAMPLE 2 Find an angle measure inside a circle Find the value of x. SOLUTION The chords JL and KM intersect inside the circle. Use Theorem 10.12. x o = 1 2 (mJM + mLK) x o = 1 2 (130 o + 156 o ) Substitute. x o = 143 Simplify.
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The chords JL and KM intersect inside the circle. ( m JM + m LK ). x o. =. 12. 12. x o. ( 130 o + 156 o ). =. x o. = 143. EXAMPLE 2. Find an angle measure inside a circle. Find the value of x. SOLUTION. Use Theorem 10.12. Substitute. Simplify. - PowerPoint PPT Presentation
Transcript of EXAMPLE 2
EXAMPLE 2 Find an angle measure inside a circle
Find the value of x.
SOLUTION
The chords JL and KM intersect inside the circle.
Use Theorem 10.12.xo = 12
(mJM + mLK)
xo = 12
(130o + 156o) Substitute.
xo = 143 Simplify.
EXAMPLE 3 Find an angle measure outside a circle
Find the value of x.
SOLUTION
Use Theorem 10.13.
Substitute.
Simplify.
The tangent CD and the secant CB intersect outside the circle.
= 12
(178o – 76o)xo
= 51 x
m BCD (mAD – mBD)= 12
EXAMPLE 4 Solve a real-world problem
SCIENCE
The Northern Lights are bright flashes of colored light between 50 and 200 miles above Earth. Suppose a flash occurs 150 miles above Earth. What is the measure of arc BD, the portion of Earth from which the flash is visible? (Earth’s radius is approximately 4000 miles.)
EXAMPLE 4 Solve a real-world problem
SOLUTION
Use Theorem 10.13.
Substitute.149o 12
[(360o – xo) –xo]
Solve for x.xo 31
= 12
m BCD (mDEB – mBD)
Because CB and CD are tangents,CB AB and CD AD
Also,BC DC and CA CA . So, ABC ADC by the Hypotenuse-Leg Congruence Theorem, and BCA DCA.Solve right CBA to find that m BCA 74.5°.
ANSWERThe measure of the arc from which the flash is visible is about 31o.
GUIDED PRACTICE for Examples 2, 3, and 4
4. Find the value of the variable.
SOLUTION
The chords AC and CD intersect inside the circle.
Use Theorem 10.12.
Substitute.
Simplify.
= 12
(yo + 95o) 78o
= y 61
78° (mAB + mCD)= 12
GUIDED PRACTICE for Examples 2, 3, and 4
Find the value of the variable.5.
SOLUTION
The tangent JF and the secant JG intersect outside the circle.
Use Theorem 10.13.
Substitute.
Simplify.
= 12
(ao – 44o) 30o
= 104 a
m FJG (mFG – mKH)= 12
GUIDED PRACTICE for Examples 2, 3, and 4
6. Find the value of the variable.
SOLUTION
Use Theorem 10.13.
Substitute.73.7o 12
[(xo) –(360 –x)o]
Solve for x.xo 253.7
= 12
m TQR (mTUR – mTR)
Because QT and QR are tangents, QR RS and QT TS
Also,TS SR and CA CA . So, QTS QRS by the Hypotenuse-Leg Congruence Theorem, and TQS RQS.Solve right QTS to find that m RQS 73.7°.
