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Simple geometry quest

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by engg.manik » Sun Oct 04, 2009 9:03 am
As the figure shows, A is the center of the circle, AB=AC=2, angle BAC is 120 degrees. What is the area of the triangle?

Is there a way to solve it woithout using trignometry
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by winnerhere » Sun Oct 04, 2009 9:14 am
The area is 1/2 (AB) (AC) Sin 120 = 1/2 * 2 * 2 sin120
= 2 sin120

=root 3
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by engg.manik » Sun Oct 04, 2009 9:20 am
winnerhere wrote:The area is 1/2 (AB) (AC) Sin 120 = 1/2 * 2 * 2 sin120
= 2 sin120

=root 3

U used trignometry ?????
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by ssmiles08 » Sun Oct 04, 2009 2:19 pm
engg.manik wrote:
winnerhere wrote:The area is 1/2 (AB) (AC) Sin 120 = 1/2 * 2 * 2 sin120
= 2 sin120

=root 3

U used trignometry ?????
You can draw a line bisecting 120 degree triangle, which make two 30-60-90 triangles.

hypotenuse is 2, base is sqrt(3)

2*base = 2sqrt(3)
height = 1

(1/2)*[sqrt(3)]*1 = sqrt(3)
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