Question

In Fig. 6.13, ZBAC = 90°, AD I BC and
ZBAD = 50°, then ZACD is
(a) 50° (b) 40° (c) 70° (d) 60°​

Answer 1

Given : ∠BAC = 90° ; ∠BAD = 50° ; ∠ACD = ?

we know the sum of angles in a triangle is 180°.

so in ΔABD

∠BAD + ∠ABD + ∠ADB = 180°

⇒ 50° + ∠ABD + 90° = 180°

(∵ AD⊥BC , ∠ADB = 90°)

⇒∠ABD  = 40°

NOW in  ΔABC ,

∠ABC + ∠BAC + ∠ACB = 180°

⇒ 40° + 90° + ∠ACB = 180°

⇒ ∠ACB = 50° = ∠ACD .

HOPE THIS HELPS YOU ✌️✌️☘️☘️.!!