Question

If PA and PB are tangents from an outside 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 ∠APB = 60°

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

In ΔPAC and ΔPBC

PA = PB(Tangents from an external point are equal)


∠APC = ∠BPC (Tangents from an external point are equally inclined to the segment joining center to that point)

PC = PC (Common)

Thus, ΔPAC is congruent to ΔPBC (By SAS congruency rule) ..........(1)

∴ AC = BC

Also ∠APB = ∠APC + ∠BPC

∠APC=$\frac{1}{2}$∠APB. {∠APC = ∠BPC}

$\frac{1}{2}$×60° = 30°

∠ACP + ∠BCP = 180°. {∠ACP =∠BCP}
∠ACP=$\frac{1}{2}$×180°

Now in right triangle ACP
sin30°=$\frac{AC}{AP}$
$\frac{1}{2}$=$\frac{AC}{10}$
AC=$\frac{10}{2}$=5

∴ AB = AC + BC = AC + AC (AC = BC)

⇒ AB = (5 + 5)cm = 10cm