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Cosine rule for two triangles:ABD and ABC cos B = a**2+c**2-b**2/2*a*c gives BC = 2. [ June 05, 2004: Message edited by: Ram Abdullah D'Souza ]

MH

Nick George
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Joined: Apr 04, 2004
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Well, that looks like the right answer, howerver, i am a bit lost as to how it is obtained. Using law of cosines, I was able to get the following three equations from the three triangles:

s=length of BD (half of BC) beta1+beta2=angle BAC

4s�=52-48cos(beta1+beta2)

s�=41-40cos(beta1)

s�=61-60cos(beta1)

after solving all three together, i found s to be something like 1.223.

can i have a little more explanation as to how you came at your answers?

Arjun Shastry
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Joined: Mar 13, 2003
Posts: 1878

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We can apply cosine rule only to two triangles.namely,ABD and ABC. Let 2x be length of BC.then BD=DC = x. For triangle,ABD, cosB = (4*4+x*x-5*5)/2*4*x-----------(1) For triangle ,ABC cosB = (4*4+(2x*2x)-6*6)/2*4*2x--------(2) Equating 1 &2 ,will give x=1-->BC=2