Question 23

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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: 

$\begin{align}  & {{(x+10)}^{\circ }}+2y+{{90}^{\circ }}+{{(100-y)}^{\circ }}=360 \\ & x+y={{160}^{\circ }} \\ & y={{160}^{\circ }}-x \\\end{align}$

maths year: