Question

In figure, AB and CD diameters of the circle. Also AC = 4 cm and $m\angle DPB={45}^o$.
(a) Find $m\angle DQB$.
(b) Find the radius of the circle.

Answer 1

(a) $m\angle DQB={180}^o-{45}^o={135}^o$
(Since the vertices of the quadrilateral DPBQ are points on a circle, its opposite angles are supplementary).
(b) Since $m\angle DPB={45}^o,m\angle DOB=2\times {45}^o={90}^o$
$m\angle AOC={90}^o$ (opposite angle of $\angle DOB$)
In $\Delta AOC,OA=OC$ (radii of the same circle)
So the angles opposite to these sides are equal.
$m\angle OAC=m\angle OCA={45}^o$
Thus, the angle of this triangle are ${45}^o,{45}^o$ and ${90}^o$. So the sides opposite these angles are in the ratio $1:1:\sqrt{2}$.
Now, AC = 4 cm, therefore
Radius $=OA=OC=\frac{4}{\sqrt{2}}=\frac{4\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}=2\sqrt{2}cm$