Q19 of 158 Page 159

In Fig. 6.9, BC = CA and ∠A = 40. Then, ∠ACD is equal to

Exterior angle is equal to sum of opposite interior angles


∠ACD = ∠A + ∠B


As ΔACB is isosceles triangle with AC = BC


Then ∠A must be equal to ∠B


∴ ∠ACD = 40° + 40° = 80°

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