113
Two tangents PA and PB are drawn from an external point P to a circle with centre O as shown in the given figure. If $\angle OAB = 15^\circ$, find the measures of each angle of $\Delta PAB$.
Show SolutionHide Solution↓
Solution
$\angle PAO = 90^\circ$ (the tangent at any point of a circle is perpendicular to the radius through the point of contact) (1/2 Mark)
$\angle PAB = 90^\circ- 15^\circ = 75^\circ$ (1/2 Mark)
As PA = PB (1 Mark)
$\therefore \angle PAB = \angle PBA =75^\circ$
$\angle PAB + \angle PBA + \angle APB = 180^\circ$
$\angle APB = 180^\circ - 150^\circ = 30^\circ$ (1 Mark)
$\angle PAO = 90^\circ$ (the tangent at any point of a circle is perpendicular to the radius through the point of contact) (1/2 Mark)
$\angle PAB = 90^\circ- 15^\circ = 75^\circ$ (1/2 Mark)
As PA = PB (1 Mark)
$\therefore \angle PAB = \angle PBA =75^\circ$
$\angle PAB + \angle PBA + \angle APB = 180^\circ$
$\angle APB = 180^\circ - 150^\circ = 30^\circ$ (1 Mark)