Question

Find the smallest angle of a cyclic quadrilateral $ABCD$ in which $\angle A=(2x-10{)}^o,\angle B=(2y-20{)}^o,\angle C=(2y+30{)}^o$ and $\angle D=(3x+10{)}^o.$

• A. ${80}^0$
• B. ${50}^0$
• C. ${85}^0$
• D. ${40}^0$
  

Answer 1

Answer: (B)

${50}^0$

Given $\angle A=(2x-10{)}^{\circ },\angle B=(2y-20{)}^{\circ },\angle C=(2y+30{)}^{\circ }$ and $\angle D=(3x+10{)}^{\circ }$.

We know, the sum of the opposite angles of a cyclic quadrilateral is ${180}^{\circ }$.

So $\angle A+\angle C={180}^{\circ }$

$\Rightarrow (2x-10+2y+30{)}^{\circ }={180}^{\circ }$

$\Rightarrow (2x+2y{)}^{\circ }={160}^{\circ }------\left(1\right)$

$\&\text{also,}\angle B+\angle D={180}^{\circ }$

$\Rightarrow (2y-20+3x+10{)}^{\circ }={180}^{\circ }$

$\Rightarrow (3x+2y{)}^{\circ }={190}^{\circ }------\left(2\right)$ .

Subtracting $\left(1\right)$ from $\left(2\right)$, we get,

$(3x-2x{)}^{\circ }={190}^{\circ }-{160}^{\circ }$

$\Rightarrow x^{\circ }={30}^{\circ }$.

Using $x^{\circ }={30}^{\circ }$ in $\left(1\right)$, we get,

$(2y+2\times 30{)}^{\circ }={160}^{\circ }$

$\Rightarrow (2y+60{)}^{\circ }={160}^{\circ }$

$\Rightarrow (2y{)}^{\circ }={100}^{\circ }$

$\Rightarrow y^{\circ }={50}^{\circ }$.

So, $\angle A=(2x-10{)}^{\circ }=(2\times 30-10{)}^{\circ }=(60-10{)}^{\circ }={50}^o$

$\angle B=(2y-20{)}^{\circ }=(2\times 50-20{)}^{\circ }=(100-20{)}^{\circ }={80}^o$

$\angle C=(2y+30{)}^{\circ }=(2\times 50+30{)}^{\circ }=(100+30{)}^{\circ }={130}^o$

and $\angle D=(3x+10{)}^{\circ }=(3\times 30+10{)}^{\circ }=(90+10{)}^{\circ }={100}^o$.

Hence, the smallest angle is $\angle A={50}^o$.

Therefore, option $B$ is correct.