Question

From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle, show that $△APB$ is equilateral.

Answer 1

$OA=OB=r$

$OP=2r$

In $\Delta OAP$ it is right angled at $A$

$O{A}^2+A{P}^2=O{P}^2$

$A{P}^2=O{P}^2-O{A}^2=9r^2-r^2=3r^2$

$AP=\sqrt{3}r$

Similarly $BP=\sqrt{3}r$

In $\Delta OAP,\tan \theta =\frac{r}{\sqrt{3}r}=\frac{1}{\sqrt{3}}\to \theta ={30}^{\circ }$

$\alpha ={90}^{\circ }-{30}^{\circ }={60}^{\circ }$

In $\Delta OAT$

$\sin \alpha =\frac{AT}{r}$ $\frac{\sqrt{3}}{2}=\frac{AT}{r}$

$AT=\frac{\sqrt{3}}{2}r$

$AT=BT=\frac{\sqrt{3}}{2}r$

$AB=\sqrt{3}r$

In $\Delta APB$ $AP=AB=BP=\sqrt{3}r$

Hence it is equilateral.