In a triangle ABC, ∠B = 50° and ∠A = 60°. Suppose BC is extended to D. Find ∠ACD.

It is given that in ∆ ABC,
∠B = 50° and ∠A = 60°
Here, ∠ACD is an external angle. We know that an exterior angle of a triangle is always larger than either of the interior opposite angles.
⇒ ∠ACD = ∠A + ∠B = 50° + 60° = 110°
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