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Require assistance - how to solve this

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Re: Require assistance - how to solve this [#permalink] New post 07 Jun 2011, 09:45
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arryputtar wrote:
One of the angles of a triangle is 60° and length of its two sides is 6 cm and 7 cm. The length of the
third side of the triangle is
a. \sqrt{43} b. 3+\sqrt{22} + c. \sqrt{43} or 3 +\sqrt{22} d. None of these


I dont think this is tested on GMAT

formula is
a^2= b^2+C^2-2bc Cos A
a^2 = 36 + 49- 2*6*7*cos 60
a^2 = 36 + 49- 2*6*7*1/2
a^2 = 43
a = sqr 43.

hence A
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Re: Require assistance - how to solve this [#permalink] New post 10 Jun 2011, 21:49
agree with sudhir. this is not tested on the GMAT.
If it was mentioned that its a right angled triangle etc then it would make sense.
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Re: Require assistance - how to solve this [#permalink] New post 11 Jun 2011, 03:17
Thats a bit too formula-centric..dont think it would be tested..
Agree with the solution provided..
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Re: Require assistance - how to solve this [#permalink] New post 14 Jun 2011, 00:33
well the above formula is derived.
Drop a perpendicular from A on to BC and name the point D.

now in triangle CDA, cos 60 = CD/AC = 1/2
CD = 3

thus AD^2 = 6^2-3^2 = 27
now in triangle BDA,
AB^2 = AD^2 + BD^2
BD = 7-3 = 4

thus AB^2 = 16 + 27 = 43.Hence.
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Re: Require assistance - how to solve this   [#permalink] 14 Jun 2011, 00:33
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