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Geometry

by jsasipriya » Sun Jan 02, 2011 1:31 am
A regular pentagon is inscribed in a circle. If A and B are adjacent vertices of the pentagon and O is the center of the circle, what is the value of angle OAB ?

A 48 degrees
B 54 degrees
C 72 degrees
D 84 degrees
E 108 degrees

OA B

How to solve this? Please help.
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by Anurag@Gurome » Sun Jan 02, 2011 3:00 am
jsasipriya wrote:A regular pentagon is inscribed in a circle. If A and B are adjacent vertices of the pentagon and O is the center of the circle, what is the value of angle OAB ?

A 48 degrees
B 54 degrees
C 72 degrees
D 84 degrees
E 108 degrees
Image

Refer to the above image.

Note that angle AOB = One-fifth of 360° = 360°/5 = 72°
Also note that the triangle OAB is isosceles with OA and Ob as equal sides.
Thus angle OAB = angle OBA

Now, Angle OAB + Angle OBA + Angle AOB = 180°
=> Angle OAB + Angle OAB + 72° = 180°
=> 2*(Angle OAB) = (180° - 72°) = 108°
=> Angle OAB = 108°/2 = 54°

The correct answer is B.
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by ragz » Sun Jan 02, 2011 7:07 pm
jsasipriya wrote:A regular pentagon is inscribed in a circle. If A and B are adjacent vertices of the pentagon and O is the center of the circle, what is the value of angle OAB ?

A 48 degrees
B 54 degrees
C 72 degrees
D 84 degrees
E 108 degrees

OA B

How to solve this? Please help.
It's a general rule that the sum of internal angles of a polygon is (n-2)*180, where n is the number of sides of a polygon. Since, the polygon in question is a pentagon, the sum of interior angles is 540 degrees. Also, since it is a regular polygon, each angle is equal to 540/5 = 108 degrees.

The line from Center to the vertice of the pentagon will bisect the angle of the pentagon into half. Hence, the angle in question is 108/2 = 54 degrees.

Hence, answer is B.
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