Question

The angles of a cyclic quadrilateral $ABCD$ are $A=(6x+10)$, $B=\left(5x\right)$, $C=(x+y)$, $D=(3y-10)$.
Find $x$ and $y$, and hence find the values of the four angles.

• A. $x=10,y=30,A={120}^o,B={100}^o,C=506O,D=806o$
• B. $x=20,y=30,A={130}^o,B={100}^o,C={50}^o,D={80}^o$
• C. $x=18,y=30,A={140}^o,B={100}^o,C={50}^o,D={84}^o$
• D. $x=30,y=30,A={180}^o,B={100}^o,C={50}^o,D={80}^o$

Answer 1

The correct option is B $x=20,y=30,A={130}^o,B={100}^o,C={50}^o,D={80}^o$

As $ABCD$ is a cyclic quadrilateral, we have

$\angle A+\angle C={180}^o$ .......(The sum of the opposite angles of a cyclic quadrilateral $={180}^o$)

$⟹$ $6x+{10}^o+x+y={180}^o$

$⟹$$7x+y={170}^o$ .....(1)

and $\angle B+\angle D={180}^o$ ............(The sum of the opposite angles of a cyclic quadrilateral $={180}^o$)

$⟹$$(3y-10)+5x={180}^o$

$⟹$$3y+5x={190}^o$ ....(2).

Solving (1) and (2), we have,

$x={20}^o$

$y={30}^o$

$A=6x+10=6\times {20}^o+{10}^o={130}^o$

$B=5x=5\times {20}^o={100}^o$

$C=x+y={30}^o+{20}^o={50}^o$

$D=3y-10=3\times 30-10={80}^o$.

Therefore, option $B$ is correct.