Question

Use the information given in the following figure to find:
(1) x
(2) $\angle B$ and $\angle C$

Answer 1

Here, given that,
$\angle A={90}^{\circ }\angle B=(2x+4{)}^{\circ }\angle C=(3x-5{)}^{\circ }\angle D=(8x-15{)}^{\circ }$
We know that,
Sum of all the interior angles in a quadrilateral is ${360}^{\circ }$
So,
$\angle A+\angle B+\angle C+\angle D={360}^{\circ }$
${90}^{\circ }+(2x+4^{\circ })+(3x-5^{\circ })+(8x-{15}^{\circ })={360}^{\circ }$
On further calculation, we get
${90}^{\circ }+2x+4^{\circ }+3x-5^{\circ }+8x-{15}^{\circ }={360}^{\circ }{74}^{\circ }+13x={360}^{\circ }13x={360}^{\circ }-{74}^{\circ }13x={286}^{\circ }x={\left(\frac{286}{13}\right)}^{\circ }\text{We get,}x={22}^{\circ }\text{The value of}xis{22}^{\circ }\text{Now,}\angle B=2x+4^{\circ }=2\times {22}^{\circ }+4^{\circ }={48}^{\circ }\angle C=3x-5^{\circ }=3\times {22}^{\circ }-5^{\circ }={61}^{\circ }$
Therefore, $\angle B={48}^{\circ }$ and $\angle C={61}^{\circ }$