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In ∆PQR 3sinP+4cosQ=6 and 4sinQ +3cosP=1 then the angle R is equal to?

In ∆PQR 3sinP+4cosQ=6 and 4sinQ +3cosP=1 then the angle R is equal to?

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Grade:11

1 Answers

Arun
25750 Points
6 years ago
square both the equations and add them-
we get-
9(sin2p + cos2p) + 16(sin2q + cos2q) + 24(sinq cosp + cosq sinp) = 36 +1
9 + 16 + 24 sin (p + q) = 37
sin (p + q) = ½
p + q = 30 degree
hence angle R will be = 180 –30
 = 150 degree = 5\pie\pi / 6 radian
hope it helps

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