Question

In fig,O is the centre of the circle. The angle made by the arc BCD at centre is 150°.BC is produced to P. Find angle DCP

Answer 1

very simple !!

<BAD = 1/2 <BOD

<BAD = 1/2 × 150°

<BAD = 75°

now , from the exterior angle property

< BAD = < DCP

=> 75°

hope it help you !!

thanks !!

Answer 2

• ∠BAD = Half of ∠BOD       [ Angle suspended by an arc on center is double that of the circumference]

 So ∠ BAD = 150/2 = 75°

• ∠BCD = 180 - 75  [ Sum of the oppo. angle of a cyclic  Quadrilateral is 180 ]

So   ∠ BCD = 105°

• So ∠DCP = 180 - 105 = 75°.   [ Linear pair].

thank u.