Question

If PA and PB are tangents drawn from external point P such that PA=$10$cm and $\angle APB={60}^{\circ }$ find the length of chord AB?

Answer 1

Given PA and PB are tangents of a circle, PA= 10 cm and $\angle APB={60}^{\circ }$

Let O be the center of the given circle and C be the point of intersection of OP and AB.

In $\Delta PAC$ and $\Delta PBC$

PA=PB (Tangents from an external points are equally inclined to the segment joining center to that point)

PC=PC(Common)

Thus, $\Delta APC$ is congruent to $\Delta PBC$ (By SAS congruency rule) _______ (1)

Also $\angle APB=\angle APC+\angle BPC$

$\angle APC=\frac{1}{2}.\angle PAB[\because \angle APC=\angle BPC]$

$\frac{1}{2}\times {60}^{\circ }={30}^{\circ }$

$\angle ACP+\angle BCP={180}^{\circ }[\angle ACP=\angle BCP=\angle ACP=\frac{1}{2}\times {180}^{\circ }]$

Now in right triangle ACP

$sin{30}^{\circ }=\frac{AC}{AP}$

$\frac{1}{2}=\frac{AC}{10}$

$AC=\frac{10}{2}$

$AC=5$

$\therefore AB+AC+BC=AC+AC(AC=BC)$

$\therefore$ The length of the chord AB is 10 cm.