waecmaths question:
The angle of a quadrilateral are ${{(x+10)}^{o}},2{{y}^{o}},{{90}^{o}}\text{ and (100}-y{{)}^{o}}$. Find y in term of x
Option A:
$y=160+x$
Option B:
$y=100+x$
Option C:
$y=160-x$
Option D:
$y=x-100$
waecmaths solution:
$\begin{align} & {{(x+10)}^{\circ }}+2y+{{90}^{\circ }}+{{(100-y)}^{\circ }}=360 \\ & x+y={{160}^{\circ }} \\ & y={{160}^{\circ }}-x \\\end{align}$
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