thanks rahul mishra.
that was a better straight forward way. :-)
In triangle ABC, we are given that 3sin{A}+4cos{B}=6 and 4sin{B}+3cos{A}=1 then find angle C.
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7 Answers
h4hemang
·2011-11-03 22:13:58
take 4cosB to RHS similarly take 4sinB to RHS.
square both the sides of both the equations and add.
this will cancel out cosA and sinA.
Shubhodip
·2011-11-04 10:32:40
he is sharing problems and its solutions
its boring to wait for others
no problems!!
:)
rahul
·2011-11-04 13:31:20
Simply square and add both the eqns.
i.e., 9 sin2A + 16 cos2B + 24 sinA.cosB + 16 sin2B + 9 cos2A + 24 cosA.sinB = 37
=> 9 + 16 + 24 sin (A + B) = 37
=> sin(A + B) = 12/24
=> sin C = 1/2 or C = 30°