waecmaths question:
In the diagram, PR is a diameter, $\angle PRQ={{(3x-8)}^{\circ }}$and$\angle RPQ={{(2y-7)}^{\circ }}$. Express x in terms of y
Option A:
$x=\frac{75-2y}{3}$
Option B:
$x=\frac{105-3y}{2}$
Option C:
$x=\frac{105-2y}{3}$
Option D:
$\frac{75-3y}{2}$
waecmaths solution:
$\begin{align} & \angle PQR={{90}^{\circ }}\text{ }\!\!\{\!\!\text{ Angle in a semicircle }\!\!\}\!\!\text{ } \\ & \therefore {{(3x-8)}^{\circ }}+{{(2y-7)}^{\circ }}+{{90}^{\circ }}={{180}^{\circ }} \\ & {{(3x-8)}^{\circ }}+{{(2y-7)}^{\circ }}={{90}^{\circ }} \\ & 3x+2y=105 \\ & 3x=105-2y \\ & x=\frac{105-2y}{3} \\\end{align}$
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