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Question 13

In triangle $$PQR$$, $$\angle R=2\angle P$$, $$PR=5$$ and $$QR=4$$. The length of $$PQ$$ is

Solution

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Let $$PQ=x$$ and $$\angle P=\theta$$. The sine rule gives $$\frac{x}{\sin2\theta}=\frac{4}{\sin\theta}$$, so $$x=8\cos\theta$$. Applying the cosine rule at $$P$$ gives $$16=25+x^2-10x\cos\theta=25-\frac{x^2}{4}$$, and hence $$x=6$$.

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