In the given figure, the diameter AB of the circle with centre O is extended to a point P and PQ is a tangent to the circle at the point T. If BPT = x and ATP = y, then prove that x + 2y = 90o.

Asked by Topperlearning User | 27th Jul, 2017, 01:04: PM

Expert Answer:

Join OT.

It is known that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

OTP = 90o

In a triangle, the measure of an exterior angle is equal to the sum of the measures of its interior opposite angles.

Therefore, in PAT,

OAT = APT + ATP

OAT = x + y

In OAT, OA = OT (Radii of the same circle)

OAT = OTA

x + y = OTP - ATP

x + y = 90o - y

x + 2y = 90o

Answered by  | 27th Jul, 2017, 03:04: PM

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