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Area of triangle, given ratio of interior angles [#permalink]
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01 Nov 2009, 04:06
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Re: Area of triangle, given ratio of interior angles [#permalink]
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01 Nov 2009, 04:56



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Re: Area of triangle, given ratio of interior angles [#permalink]
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01 Nov 2009, 04:19
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angles in ratio x:2x:3x so its a 306090 triangle and with shortest side 1, from trigonometry we have hypotenuse =2 and longest side =\sqrt{3},
so area =1/2 * base * height = 1/2 * 1 *\sqrt{3} = \sqrt{3}/2 which is D



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Re: Area of triangle, given ratio of interior angles [#permalink]
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01 Nov 2009, 04:22
x+2x+3x=180
x=180/6, x=30
so the angles are 30,60,90
and AB is the shortest side and it has to be oppsite to 30.
in 306090 combination the ratio of lengths are 1:\sqrt{3}:2
so the one opp to 60 is \sqrt{3}
one opp to 90 is 2
the area is \sqrt{3}



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Re: Area of triangle, given ratio of interior angles [#permalink]
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01 Nov 2009, 04:41
divyakatas wrote: x+2x+3x=180
x=180/6, x=30
so the angles are 30,60,90
and AB is the shortest side and it has to be oppsite to 30.
in 306090 combination the ratio of lengths are 1:\sqrt{3}:2
so the one opp to 60 is \sqrt{3}
one opp to 90 is 2
the area is \sqrt{3} kirankp has it right and you are very close, something wrong in the final step... OA:



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Re: Area of triangle, given ratio of interior angles [#permalink]
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09 Jul 2014, 06:39
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Re: Area of triangle, given ratio of interior angles [#permalink]
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18 Jul 2016, 19:02
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Re: Area of triangle, given ratio of interior angles
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