Question

AB is a chord of a circle with centre O. If $\angle AOB={50}^{\circ }$ and AB = 18 cm, then find the diameter of the circle $sin{25}^{\circ }=0.42$.

Answer 1

Draw diameter AC passing through center O.

$\angle BOC=180-{50}^{\circ }={130}^{\circ }$

$\angle OBC=\angle OCB=\frac{({180}^{\circ }-{130}^{\circ })}{2}={25}^{\circ }$ $\left[1mark\right]$

In right-angled triangle ABC

$sin{25}^{\circ }=\frac{AB}{AC}$

$AC=\frac{AB}{sin{25}^{\circ }}=\frac{18}{0.42}=42.85$

So, diameter $AC=42.85cm$ $\left[1mark\right]$