Question

In a circle with centre o ab is diameter and points c,d,e are on the one side of a b such that abcd is a pentagon find angle acd+angle deb

Answer 1

The value of ∠ACD + ∠BED is 270°

Centre of the circle = O (Given)

Diameter of the circle = AB (Given)

Pentagon = ABCD (Given)

Construction - Join AE,

As ACDE is a cyclic quadrilateral,

Therefore,

∠ACD + ∠AED = 180° --- 1

∠AEB = 90° ( Since, the angle is in semi circle) --- 2

Adding equation 1 and 2 -

∠ACD + ∠AED + ∠AEB  = 180° + 90°  

∠ACD + ∠BED = 270°  

Therefore, the value of ∠ACD + ∠BED is 270°