In a circle I is the incentre of triangle ABC, BI  when produced meets the circumcircle of triangle ANC at D. 

Given that Angle BAC=55 degrees; Angle ACB= 65degrees

Calculate:-

i) Angle DCA

ii)Angle DAC

iii) Angle DCI

iv) Angle AIC

Dear Student!

Here is the answer to your query.

Given : ∠BAC = 55° as ∠ACB = 65°

I is the incentre of triangle ABC

⇒ AI, BI and CI are the angle bisectors of ∠BAC, ∠ABC and ∠ACB

Now in ΔABC

∠BAC + ∠ABC + ∠ACB = 180°  (Angle sum property)

⇒ 55° + ∠ABC + 65° = 180°

⇒ ∠ABC + 120° = 180°

⇒ ∠ABC = 180° – 120° = 60°

We know that angles in the same segment of the circle are equal.

⇒ ∠DCA = ∠ABD = ∠ABC =

∠DCA = ∠DBC = ∠ABC =

Now ∠DCI = ∠DCA + ∠ACI = ∠BCA

In ΔACI

∠CAI + ∠ACI + ∠AIC = 180° (Angle sum property)

⇒ ∠BAC + ∠BCA + ∠AIC = 180°

Cheers!