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The figure shows that (BP) is a line passing the center O and (PT) is

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MathRevolution
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The figure shows that (BP) is a line passing the center O and (PT) is [#permalink] New post   Updated on: 14 Aug 2020, 01:27
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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 = \(20^o\), what is ∠x?

Attachment:
PS Circle & Triangle .jpg
PS Circle & Triangle .jpg [ 7.42 KiB | Viewed 1689 times ]


A. \(40^o\)
B. \(45^o\)
C. \(50^o\)
D. \(55^o\)
E. \(60^o\)
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D

Originally posted by MathRevolution on 14 Aug 2020, 01:21.
Last edited by Bunuel on 14 Aug 2020, 01:27, edited 1 time in total.
Renamed the topic.
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Re: The figure shows that (BP) is a line passing the center O and (PT) is [#permalink] New post   14 Aug 2020, 03:55
MathRevolution wrote:
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 = \(20^o\), what is ∠x?

A. \(40^o\)
B. \(45^o\)
C. \(50^o\)
D. \(55^o\)
E. \(60^o\)


tangent PT makes right angle with radius OT.
In triangle POT, ∠POT = \(180 - (20 + 90) = 70^o\)

In triangle AOT, AO and OT are radius and thus ∠OAT = ∠OTA = x
\(x + x + ∠AOT = 180\)
\(2x = 180 - 70\)
\(x = 55^o\)
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The figure shows that (BP) is a line passing the center O and (PT) is [#permalink] New post   Updated on: 17 Sep 2020, 04:46
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Solution:

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

Since the triangle is an isosceles triangle with AO = OT, we have ∠OTA = ∠OAT = ∠x.

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

Therefore, D is the correct answer.

Answer: D

Originally posted by MathRevolution on 16 Aug 2020, 01:55.
Last edited by MathRevolution on 17 Sep 2020, 04:46, edited 1 time in total.
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Re: The figure shows that (BP) is a line passing the center O and (PT) is [#permalink] New post   04 Sep 2020, 05:20
MathRevolution wrote:
Solution:

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

Since the triangle is an isosceles triangle with AO = OT, we have ∠OTA = ∠OAT = ∠x.

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

Therefore, D is the correct answer.

Answer: D


Is there a rule why the circle must also go through a and build an isoceles traingle? I mean just because the drawing looks like it is not enough to confirm that it actually does?
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Re: The figure shows that (BP) is a line passing the center O and (PT) is [#permalink]
04 Sep 2020, 05:20
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