Hi there, Can anyone give me help to start off this problem please? Triangle PQR Angle PRQ = 3x; Angle PQR = x I have to prove that p = 4q cos x cos 2x. Thanks Cromlix
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Originally Posted by cromlix Hi there, Can anyone give me help to start off this problem please? Triangle PQR Angle PRQ = 3x; Angle PQR = x I have to prove that p = 4q cos x cos 2x. Thanks Cromlix law of sines ... $\displaystyle \dfrac{q}{\sin{x}} = \dfrac{p}{\sin(180-4x)}$ $\displaystyle q\sin(180-4x) = p\sin{x}$ ... taking it from this point will lead to what you want proved
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