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In triangle ABC, we are given that 3sin{A}+4cos{B}=6 and 4sin{B}+3cos{A}=1 then find angle C.

#Mathematics #Algebra

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7 Answers

3

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.

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71

Vivek @ Born this Way ·2011-11-04 09:27:48

You give , You Solve?

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Shubhodip ·2011-11-04 10:32:40

he is sharing problems and its solutions

its boring to wait for others

no problems!!

:)

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71

Vivek @ Born this Way ·2011-11-04 10:35:49

Wow!

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36

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 = 30Â°

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h4hemang ·2011-11-05 07:23:48

thanks rahul mishra.
that was a better straight forward way. :-)

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ayush_2008 ·2011-11-05 20:55:46

but about 150

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