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Question on Geometry

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by rahdesh » Tue Feb 02, 2010 10:37 am
This is an easy one, but i dont know why i got 45 as my answer the first time. can neone please tell me.


If A is the center of the circle shown above and AB=BC=CD, what is the value of x?
(A) 15
(B) 30
(C) 45
(D) 60
(E) 75

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by Osirus@VeritasPrep » Tue Feb 02, 2010 10:47 am
since AB= BC = CD we can now draw in two equilateral triangles. An equality triangle has 3 angles that are 60 degrees each. Since we know that angle BAC is 60 degrees, and we know the other angle is 90 degrees, then we have a 30-60-90 right triangle. X therefore equals 30.
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by thephoenix » Tue Feb 02, 2010 10:56 am
in triangle ABC
ab=bc
ab=ac---radius

therefore ab=bc=ac--->triangle ABC is equilateral

same is true for tri ADC

therefor <BAC=60
AND < DAC=60

<BAD=BAC+DAC=120
IN TRI ABD
BAD+2X=180
--->X=(180-120)/2=30
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by ajith » Tue Feb 02, 2010 10:59 am
In Triangle ABC; BC=AB=AC(radius) so ABC is an equilateral and hence angle BAC =60 degree
In Triangle ACD; CD=AC=AD(radius) so ACD is an equilateral and hence angle CAD =60 degree

angle BAD = BAC+CAD =120 degree

In triangle BAD AB = AD so ABD is an isosceles and ABD =BDA =x

now sum of all interior angles of triangle BAD = 180

BAD + ABD + BDA = 180
120+2x =180
x=30
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by harshavardhanc » Tue Feb 02, 2010 12:33 pm
osirus0830 wrote:since AB= BC = CD we can now draw in two equilateral triangles. An equality triangle has 3 angles that are 60 degrees each. Since we know that angle BAC is 60 degrees, and we know the other angle is 90 degrees, then we have a 30-60-90 right triangle. X therefore equals 30.
the quickest way!

others too have posted a way, but this one's really fast!

here's another one :

<ADB = x

=> <BAD = 180 - 2x

=> <BCD = 90 - x ( angle subtended by the chord on the circle is half the angle subtended by it at the center)

and as <BAD + <BCD = 180 (sum of oppo angles of a quadrilateral)

180 - 2x + 90 - x = 180

=> x = 30

I know this definitely looks lengthier , but the calculations are oral if you know the property.
Regards,
Harsha
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