Question

someone please answer my question
this is the second time i am asking, but no reply -

Aob is a diameter of the circle with centre o. The tangent at point T on the circle meets AB produced at P. If BAT = 30 degree, find angle TPA
Experts please answer who knows

Answer 1

Hope this helps u :-)

Answer 2

Here is your answer mate !!

Consider Δ OTA

= OT = OA ( Radius )

Therefore OTA is an isosceles triangle.

⇒ ∠ OAT = ∠ OTA = 30°

We know that Radius is perpendicular to tangent of a circle. Hence,

= ∠ OTP = 90°

= ∠ ATP = ∠ OTA + ∠ OTP

= ∠ ATP = 30° + 90°  = 120°  

Now consider Δ ATP

= ∠ ATP + ∠ PTA + ∠ PAT = 180°  ( Angle sum property of a triangle )

= 120° + 30° + ∠ PAT = 180°

= 150° + ∠ PAT = 180°

= ∠ PAT = 180° - 150°

⇒ ∠ PAT = 30°

Hope this helps !!

Cheers !!