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 40 Page 766

Recall that the angle formed is half the half the measure of the difference of the intercepted arcs.

See solution.

Practice makes perfect
Let's begin by drawing a circle centered at P.
Next, let's draw two lines tangent to the circle.

With the help of a protractor we will find the measure of ∠ Q.

As we can see, m∠ Q = 50^(∘). Now, by the Arc Addition Postulate, we can write the following equation and solve if for one of the measures. mACB + mAB = 360^(∘) ⇓ mAB = 360^(∘) - mACB Now, since the two tangent lines intersect each other outside the circle, the measure of ∠ Q is half the measure of the difference of the intercepted arcs. m∠ Q = 1/2(mACB - mAB) Let's substitute the expression we found for mAB and m∠ Q into the equation above.
m∠ Q = 1/2(mACB - mAB)
50^(∘) = 1/2(mACB - ( 360^(∘)-mACB))
Solve for mACB
50^(∘) = 1/2(mACB - 360^(∘)+mACB)
50^(∘) = 1/2(2mACB - 360^(∘))
100^(∘) = 2mACB - 360^(∘)
460^(∘) = 2mACB
230^(∘) = mACB
mACB = 230^(∘)
With this measure, let's find the measure of the minor arc. mAB = 360^(∘) - mACB^(230^(∘)) ⇓ mAB = 130^(∘) To summarize, let's mark all the measures in our diagram.