Question 39

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Jambmaths question: 

If the area of $\vartriangle PQR$above is $12\sqrt{3}$cm2, find the value of q?

Jamb Maths Solution: 

$\begin{align}  & A=\tfrac{1}{2}rq\sin \theta  \\ & 12\sqrt{3}=\tfrac{1}{2}\times 8\times q\sin 6{{0}^{\circ }} \\ & 12\sqrt{3}=4q\times \frac{\sqrt{3}}{2} \\ & q=6cm \\\end{align}$

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