The area of the $\triangle PQR$ is equal to $48.21cm^2$ if $PR = 15cm$ and $\angle PRQ = 40°$

I know we use the sine rule by how do we use it wit 1 angle and one side?

What I did is $15\sin(40) ÷ 48.21 = 0.1999$

$\sin^{-1} (0.1999) = 11.53$

My answer booklet said the answer is $10$

How did they get this answer?


You can get two equations: $$A=\frac{pr}{2}\sin(40^{\circ})$$ and $$15^2=p^2+r^2-2pr\cos(40^{\circ})=p^2+r^2-4A\cot(40^{\circ})$$(Note that $$pr=\frac{2A}{\sin(40^{\circ})}$$). With these equations you can compute $p=RQ$ or $r=PQ$.


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