Question

In the given figure, if $y={36}^{\circ }$ and $z={40}^{\circ }$ determine x. If y and z were complementary angles, show that x would have been $\frac{1}{2}$ right angle.

Answer 1

It is clear that APCP is a cyclic quadrilateral
$\therefore x^{\circ }$ is external angle to this cyclic quadrilateral.
$\Rightarrow$ x= internally opposite angle $=\angle EAC$
$\therefore \angle EAC=x^{\circ }$
Now, $\angle PCB$ is external angle to $\Delta PCD$
$\therefore \angle PCB=x^{\circ }+y^{\circ }$
Considering $\Delta EBC$
$\angle BEC+\angle ECB+\angle EBC={180}^{\circ }$
$z^{\circ }+x^{\circ }+y^{\circ }+x^{\circ }={180}^{\circ }$
or $x=\frac{180-(y+z)}{2}$
Given $y={36}^{\circ }$ and $z^{\circ }={40}^{\circ }$
(i) $x=\frac{180-(36+40)}{2}=90-38={52}^{\circ }$ (1)
(ii) If $y+z={90}^{\circ }$, then $x=\frac{180-90}{2}=\frac{1}{2}\times {90}^{\circ }$
$=\frac{1}{2}$ right angle (2)