Question

If the angle between two tangents drawn from an external point p to a circle of radius a and centre o , is 60 degree then find the lenght of op.

Answer 1

Step-by-step explanation:

Let the tangents from external point P to the cirle with centre O touches the circle at points A and B.

OA and OB are radii.

Therefore, OA = OB = a

$\because$ tangent is perpendicular to the radius.

$\therefore OA\perp PA \: \& \:OB\perp PB$

$\therefore \angle PAO =\angle PBO=90\degree$

$\because$ OP is the bisector of $\angle APB\\</p><p>\therefore\angle APO =\angle BPO=\frac {1}{2}\times \angle APB\\</p><p>\therefore\angle APO =\angle BPO=\frac {1}{2}\times 60\degree \\</p><p>\therefore\angle APO =\angle BPO=30\degree\\ \therefore \angle AOP =\angle BOP</p><p>\\= 180\degree - (90\degree +30\degree)\\</p><p>\therefore \angle AOP =\angle BOP= 180\degree - 120\degree \\</p><p>\therefore \angle AOP =\angle BOP= 60\degree$

$\therefore \triangle APO\: \&\: \triangle BPO$is of measure $30\degree, 60\degree \:\&\: 90\degree\: \triangle$

In $\triangle APO \:\&\: \triangle BPO,$ PO is hypotenuse and radii OA and OB are opposite to $30\degree$ angle.

$\because$ side opposite to $30\degree$ is half of hypotenuse.

$\therefore OA = OB = \frac{1}{2} \times OP\\</p><p>\therefore a= \frac{1}{2} \times OP\\</p><p>\huge \fbox {\therefore OP = 2a}$