Question

In quadrilateral $ABCD$, if $\angle A={60}^{\circ }$ and $\angle B:\angle C:\angle D=2:3:7$, then find $\angle D$.

• A. ${175}^{\circ }$
• B. ${135}^{\circ }$
• C. ${150}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

Answer: (A)

${175}^{\circ }$

In quadrilateral $ABCD$, $\angle A$ = ${60}^{\circ }$ and $\angle B:\angle C:\angle D=2:3:7$ [Given].

Let $\angle B,\angle C,\angle D$ are $2x^{\circ }$ , $3x^{\circ }$ and $7x^{\circ }$.

$\Rightarrow$ $\angle A+\angle B+\angle C+\angle D$=${360}^{\circ }$ [Sum of all angles of quadrilateral is $360$${}^{\circ }$]

$\Rightarrow$ ${60}^{\circ }+2x^{\circ }+3x^{\circ }+7x^{\circ }$ = ${360}^{\circ }$

$\Rightarrow$ $12x^{\circ }$ = ${360}^{\circ }-{60}^{\circ }$

$\Rightarrow$ $x^{\circ }$ = $\frac{{300}^{\circ }}{12}$

$\Rightarrow$ $x^{\circ }$ = ${25}^{\circ }$.

Now, $\angle D$=$7x^{\circ }$.

Substitute value of $x$ we get,

$\Rightarrow$ $\angle D$=$7\times {25}^o$

$\therefore$ $\angle D={175}^{\circ }$.

Hence, option $A$ is correct.