Question

In the adjoining figure, I is the incentre of ΔАBC. BI produced meets the circumcircle of ΔABC at D. If ∠BAC = 50° and ∠ACB = 70°, calculate: 

1. ∠DCA 
2. ∠DAC 
3. ∠DCI
4. ∠AIC

Answer 1

First, we find the third angle of triangle ABC using the Angle Sum Property of a Triangle (sum of interior angles = 180°).

Step 1: State the Angle Sum Property

In ΔABC:

∠BAC + ∠ABC + ∠ACB = 180°

Step 2: Substitute given values

Given ∠BAC = 50° and ∠ACB = 70°:

50° + ∠ABC + 70° = 180°

Step 3: Solve for angle ABC
120° + ∠ABC = 180°
∠ABC = 180° − 120°

The measure of ∠ABC is 60°.

(i) Calculate ∠DCA

∠DCA and ∠DBA both subtend the same arc, DA.

∴ ∠DCA = ∠DBA = 30°

(ii) Calculate ∠DAC

∠DAC and ∠DBC both subtend the same arc, DC.

∴ ∠DAC = ∠DBC = 30°

(iii) Calculate ∠DCI

∠DCI = ∠DCA + ∠ACI

= 30° + 35°

= 65°

(iv) Calculate ∠AIC

In triangle AIC:

∠IAC = `1/2 ∠BAC = 1/2` × 50° = 25°

∠ICA = `1/2` ∠ACB = 35°

Apply angle sum:

∠IAC + ∠ICA + ∠AIC = 180°
25° + 35° + ∠AIC = 180°
60° + ∠AIC = 180°
∠AIC = 120°