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McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. Secants, Tangents, and Angle Measures
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Exercise 32 Page 765

Use the Inscribed Angle Theorem and the Triangle Exterior Angle Theorem.

Statements
Reasons
1.
Tangents RS and RV
1.
Given
2.
m∠ VTS = 1/2mSWT m∠ STR = 1/2mST
2.
Inscribed Angle Theorem
3.
m∠ VTS = m∠ RST +m∠ R
3.
Triangle Exterior Angle Theorem
4.
1/2mSWT = 1/2mST +m∠ R
4.
Substitution
5.
m∠ R = 1/2mSWT -1/2mST
5.
Solving for m∠ R
6.
m∠ R= 1/2(mSWT - mST)
6.
Factor out 12
Practice makes perfect

Let's consider a circle and two tangents RS and RV.

Circle with two tangents from the same exterior point
By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of its intercepted arc. m∠ VTS = 12m SWT & (I) m∠ STR = 12m ST & (II) Notice that ∠ VTS is an exterior angle of △ STR. Then, applying the Triangle Exterior Angle Theorem, we conclude that its measure is equal to the sum of the measures of the two nonadjacent interior angles.

m∠ VTS = m∠ RST + m∠ R ⇓ m∠ R = m∠ VTS - m∠ RST Finally, we substitute the two equations found above into the last equation. That way we get the desired equation. m∠ R = 1/2m SWT - 1/2m RV ⇓ m∠ R = 1/2(m SWT - m RV)

Two-Column Proof

Given: & TangentsRSand RV Prove: & m∠ R = 12(mSWT - mRV) Let's summarize the proof we did above in the following two-column proof table.

Statements
Reasons
1.
Tangents RS and RV
1.
Given
2.
m∠ VTS = 1/2mSWT m∠ STR = 1/2mST
2.
Inscribed Angle Theorem
3.
m∠ VTS = m∠ RST +m∠ R
3.
Triangle Exterior Angle Theorem
4.
1/2mSWT = 1/2mST +m∠ R
4.
Substitution
5.
m∠ R = 1/2mSWT -1/2mST
5.
Solving for m∠ R
6.
m∠ R=1/2(mSWT - mST)
6.
Factor out 12
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Exercises 2 (Page 763)
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