Question

In Fig., ∠BAC = 90°, AD ⊥ BC and ∠BAD = 50°, then ∠ACD is

(a) 50° (b) 40° (c) 70° (d) 60°

Answer 1

(a) 50^o

From the question it is given that, ∠BAC = 90^o AD ⊥ BC and ∠BAD = 50^o

So, ∠DAC = ∠BAC – ∠BAD

= 90^o – 50^o

= 40^o

The, consider the ΔADC

From the rule of exterior angle property = ∠ADB = ∠DAC + ∠ACD

90^o = 40^o + ∠ACD

∠ACD = 90 – 40

∠ACD = 50^o