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Trignometry
27 Mar 2008 09:47:48 IST
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B meets AC at point D and Comments (5)
27 Mar 2008 20:53:57 IST
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Imagine the figure in your mind
Applying sin law in triangle DBC we have
BD/sin
= BC/sin (
-3/2) =>[BD/sin]sin (-3/2) = BC....(i)
Apply sine law in triangle ADB we have
AD =[sin/2/sin(-2)] BD......(ii)
subsitute values of BC and BD from equation (i) and (ii) in BC= BD+AD
we get,
[BD/sin]sin (-3/2) = BD + [sin/2/sin(-2)] BD
BD gets cancelled from both sides , solve for
Applying sin law in triangle DBC we have
BD/sin
= BC/sin (-3/2) =>[BD/sin]sin (-3/2) = BC....(i)Apply sine law in triangle ADB we have
AD =[sin
/2/sin(-2)] BD......(ii)subsitute values of BC and BD from equation (i) and (ii) in BC= BD+AD
we get,
[BD/sin
]sin (-3/2) = BD + [sin/2/sin(-2)] BDBD gets cancelled from both sides , solve for
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