Question

In the adjoining figure, PA and PB are tangents drawn from an external point P to a circle with centre O.
Prove that ∠APB = 2∠OAB
Figure

Or, in the adjoining figure, quadrilateral ABCD is circumscribed. If the radius of the incircle with centre O is 10 cm and AD ⊥ DC, find the value of x.
Figure

Answer 1

Given: PA and PB are the tangents to a circle with centre O from a point P outside it.
To prove: $\angle$APB = 2$\angle$OAB
Proof : Let $\angle$APB = x°
We know that the tangents to a circle from an external point are equal. So, PA = PB
Since, the angles opposite to the equal sides of a triangle are equal,
PA = PB ⇒ $\angle$PBA = $\angle$PAB

Also, the sum of the angles of a triangle is 180°.
∴ $\angle$APB + $\angle$PAB +$\angle$PBA = 180°
⇒ x° + 2$\angle$PAB = 180° [Since ∠PBA = ∠PAB]
$\Rightarrow \angle \mathrm{PAB}=\frac{1}{2}\left(180°-\mathrm{x}°\right)=\left(90°-\frac{1}{2}\mathrm{x}°\right)$
But PA is a tangent and OA is the radius of the given circle.
∴ $\angle$OAB + $\angle$PAB = 90°
$\angle \mathrm{OAB}+\left(90°-\frac{1}{2}\mathrm{x}°\right)=90°$
$\Rightarrow \angle \mathrm{OAB}=\frac{1}{2}\mathrm{x}°=\frac{1}{2}\angle \mathrm{APB}$
Hence, $\angle$APB = 2$\angle$OAB

OR
Disclaimer : In the figure side AB is given 27 cm, instead of AB , BP should be 27 cm.
Since the lengths of tangents from an external point to a circle are equal, we have:
BP = BQ = 27 cm
∴ QC = (BC - BQ) = (38 - 27) cm = 11 cm
⇒ CQ = CR = 11 cm
Join OR. Then, SORD is a rectangle.
∴ OS = DR = 10 cm
Hence, DC = (DR + RC) = (10 + 11) cm = 21 cm