Question

In a cyclic quadrilateral ABCD, it is given that $\angle A={\left(2x+4\right)}^{\circ },\angle B={\left(y+3\right)}^{\circ },\angle C={\left(2y+10\right)}^{\circ }\mathrm{and}\angle D={\left(4x-5\right)}^{\circ }$ .
Find the four angles.

Answer 1

The opposite angles of cyclic quadrilateral are supplementary, so
$$\begin{gathered} \angle A+\angle C={180}^{\circ } \\ \Rightarrow {\left(2x+4\right)}^{\circ }+{\left(2y+10\right)}^{\circ }={180}^{\circ } \\ \Rightarrow x+y={83}^{\circ }.....\left(i\right) \end{gathered}$$
And
$$\begin{gathered} \angle B+\angle D={180}^{\circ } \\ \Rightarrow {\left(y+3\right)}^{\circ }+{\left(4x-5\right)}^{\circ }={180}^{\circ } \\ \Rightarrow 4x+y={182}^{\circ }.....\left(ii\right) \end{gathered}$$
Subtracting (i) from (ii), we have
$3x=99\Rightarrow x={33}^{\circ }$
Now, substituting $x={33}^{\circ }$ in (i), we have
${33}^{\circ }+y={83}^{\circ }\Rightarrow y={83}^{\circ }-{33}^{\circ }={50}^{\circ }$
Therefore
$$\begin{gathered} \angle A={\left(2x+4\right)}^{\circ }={\left(2\times 33+4\right)}^{\circ }={70}^{\circ } \\ \angle B={\left(y+3\right)}^{\circ }={\left(50+3\right)}^{\circ }={53}^{\circ } \\ \angle C={\left(2y+10\right)}^{\circ }={\left(2\times 50+10\right)}^{\circ }={110}^{\circ } \\ \angle D={\left(4x-5\right)}^{\circ }={\left(4\times 33-5\right)}^{\circ }=132°-5^{\circ }={127}^{\circ } \end{gathered}$$
Hence, $\angle A={70}^{\circ },\angle B={53}^{\circ },\angle C={110}^{\circ }\mathrm{and}\angle D={127}^{\circ }$.