Question

The greatest angle of a cyclic quadrilateral $ABCD$ in which $\angle A=(2x-1{)}^o,\angle B=(y+5{)}^o,\angle C=(2y+15{)}^o$ and $\angle D=(4x-7{)}^o$ is:

• A. ${115}^o$
• B. ${120}^o$
• C. ${125}^o$
• D. ${130}^o$

Answer 1

The correct option is C ${125}^o$

In cyclic quadrilateral, the sum of opposite angles $={180}^o$
$⟹\angle A+\angle C={180}^o$
$⟹2x-1+2y+15={180}^o$
$⟹x+y=83$ ...(i).

Also, $\angle B+\angle D={180}^o$
$⟹4x+y=182$ ....(ii).

Then, by solving (i) and (ii),
$x=33$ and $y=50$.

Thus, $\angle A={65}^o,\angle B={55}^o,\angle C={115}^o,\angle D={125}^o$.
So, greatest angle $\angle D={125}^0$.
Hence, option $C$ is correct.