Question

From a point P 2 tangents PA andPB are drawn to a circle. If OP is equal to the diameter of the circle, Show that APBis equilateral.

Answer 1

Join OA and OB.
Now angle OAP and OBP = 90 degrees (tangent to radius)
Hence, triangles OAB and OBP are right angled triangles, whose common hypotenuse is OP.
Now, sin angle $BPO=\frac{OB}{OP}=\frac{1}{2}:\to$ angle BPO = 30 degrees.
Similarly angle APO = 30 degrees.
Therefore, angle APB = 60 degrees
In triangle APB, PA = PB (tangents from external point P are equal)

Hence, triangle APB is isosceles triangle and angle APB = 60 degrees.
$\to$ triangle APB is equilateral triangle. --- PROVED.