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(Geometry) The figure shows that (BP) is a line passing the center O and (PT) is a tangent line to the circle at point T

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by Max@Math Revolution » Fri Aug 14, 2020 12:25 am

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(Geometry) The figure shows that (BP) is a line passing the center O and (PT) is a tangent line to the circle at point T. If ∠APT = \(20^0\) , what is \(\angle x\) ?
PS Circle & Triangle .jpg
A. \(40^0\)
B. \(45^0\)
C. \(50^0\)
D. \(55^0\)
E. \(60^0\)
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Re: (Geometry) The figure shows that (BP) is a line passing the center O and (PT) is a tangent line to the circle at poi

by Max@Math Revolution » Sun Aug 16, 2020 2:01 am
Solution:

Since PT is tangent to the circle, we have \(\angle\ OTP\ =\ 90^o\ \ and\ \ \angle\ AOT\ =\ 180^o\ -\ 20^o-\ 90^o=\ 70^o\) .

Since the triangle is an isosceles triangle with AO = OT, we have \(\angle\ OTA\ =\ \angle\ OAT\ \ =\ \angle\ x\ \)

Thus \(\ \angle\ x\ =\ \frac{180^o-70^o}{2}\ =55^o\) .

Therefore, D is the correct answer.

Answer: D
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