In the given figure, ABCD is a cyclic quadrilateral in which CB = CD and TC is a tangent to the circle at C. If O is the centre of the circle and BC is produced to E such that ∠DCE = 110°, find:

(i) ∠DCT

(ii) ∠BOC

(i) From figure,

⇒ ∠DCE + ∠DCB = 180° [Linear pair]

⇒ ∠DCB = 180° - ∠DCE

⇒ ∠DCB = 180° - 110°

⇒ ∠DCB = 70°.

Given,,

CB = CD

∠BDC = ∠DBC = 55° [Angles opposite to equal sides in a triangle are equal]

Arc BC subtends ∠BOC at center and ∠BDC on the remaining part of the circle.

⇒ ∠BOC = 2(∠BDC)

⇒ ∠BOC = 2(55°) = 110°

⇒ ∠BCT = ∠BDC = 55° [Angles in alternate segments]

From figure,

⇒ ∠DCT = ∠DCB + ∠BCT

⇒ ∠DCT = 70° + 55° = 125°.

Hence, ∠DCT = 125°.

(ii) From part (i), we get :

∠BOC = 110°

Hence, ∠BOC = 110°.