Question

Given, in quadrilateral $ABCD$, $\angle C={64}^{\circ },\angle D=\angle C-8^{\circ },\angle A=5(a+2{)}^{\circ }$ and $\angle B=2(2a+7{)}^{\circ }$. Calculate $\angle A$.

• A. ${120}^o$
• B. ${130}^o$
• C. ${140}^o$
• D. ${150}^o$

Answer 1

The correct option is B ${130}^o$

Given$\angle C={64}^o,\angle D=\angle C-8^o={64}^o-8^o={56}^o,\angle A={5(a+2)}^o$ and $\angle B={2(2a+7)}^o.$

We know, by angle sum property, the sum of angles of a quadrilateral is ${360}^o$.

Then, $\angle A+\angle B+\angle C+\angle D={360}^o$

$\Rightarrow 5(a+2)+2(2a+7)+{64}^o+{56}^o={360}^o$

$\Rightarrow 5a+{10}^o+4a+{14}^o+{120}^o={360}^o$

$\Rightarrow 9a={360}^o-{144}^o$

$\Rightarrow 9a={216}^o$

$\Rightarrow a=\frac{{216}^o}{9}={24}^o$.

Therefore,

$\angle A={5(a+2)}^o$ $={5({24}^o+2)}^o$ $=5\times {26}^o={130}^o$.