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the intercepted arc

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the intercepted arc

by sanju09 » Wed Apr 04, 2012 2:36 am
P is a point outside of circle O. PAB is a secant and PD is a tangent to the circle O such that the measures of the intercepted arc AD and BD are 55° and 105° respectively. What is the measure of angle BPD?
(A) 160°
(B) 100°
(C) 80°
(D) 50°
(E) 25°



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by Anurag@Gurome » Wed Apr 04, 2012 2:56 am
sanju09 wrote:P is a point outside of circle O. PAB is a secant and PD is a tangent to the circle O such that the measures of the intercepted arc AD and BD are 55° and 105° respectively. What is the measure of angle BPD?
(A) 160°
(B) 100°
(C) 80°
(D) 50°
(E) 25°

[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]
Image

We have to find angle BPD, which is the angle formed by a tangent and a secant. The intercepted arcs are minor arcs.
Therefore, angle BPD = (1/2)[measure of arc BD - measure of arc AD] = (1/2)[105º - 55º] = [spoiler]25º[/spoiler]

The correct answer is E.
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by neelgandham » Wed Apr 04, 2012 3:05 am
From the attachment, [spoiler](B)100°[/spoiler]
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by Pharo » Wed Apr 04, 2012 3:35 am
What a good question! Thank you Mr. Sanjeev :)
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by sanju09 » Wed Apr 04, 2012 3:39 am
neelgandham wrote:From the attachment, [spoiler](B)100°[/spoiler]
Your attachment is not representing the question exactly. You should have treated the minor arc as the intercepted arc. [spoiler]B is not correct[/spoiler].
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by sanju09 » Wed Apr 04, 2012 3:45 am
Pharo wrote:What a good question! Thank you Mr. Sanjeev :)
That's really so sweet of you.
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