Find the Angle Between Two Tangents of a Circle

Geometry math problem: how to find the angle between two tangents

The figure shows a square ABCD, and draw a circle with a radius of half of the sides of the square and the centre of the square is at the corner of the square. PD and PB are tangents of the circle, then find the angle between the tangents of the circle. (Find ∠P)

Solution

We can connect the centre of the circle to the tangent of the circle (tangent and circle meeting point)

Let the sides of the square be 2a then the radius of the circle is a

Now from triangle BEC

sin (∠CBE) = CE/CB = a/2a = 1/2

∠CBE = 30°

So

∠BCE = 90 – 30 = 60°

Similarly ∠FCD = 60°

From this figure

∠FCE = 360 – (∠FCD + ∠BCE + ∠BCD)

⇒ ∠FCE = (60+60+90) = 150°

So

∠P = 360 – ( 150+90+90) = 30°

The angle between the two tangents is 30°