Question

$Ina\Delta ABC,\text{it is given that}\angle A:\angle B:\angle C=3:2:1and\angle ACD={90}^{\circ }.\text{If BC is produced to E then}\angle ECD=?$

$\left(a\right){60}^{\circ }\left(b\right){50}^{\circ }\left(c\right){40}^{\circ }\left(d\right){25}^{\circ }$

Answer 1

ANSWER:
Let ∠A=3x°, ∠B=2x° and ∠C=x°
Then,
3x+2x+x=180° Sum of the angles of a triangle⇒6x=180°⇒x=30°
Hence, the angles are
∠A=3×30°=90°, ∠B=2×30°=60° and ∠C=30°
Side BC of triangle ABC is produced to E.
∴∠ACE=∠A+∠B⇒∠ACD+∠ECD=90°+60°⇒90°+∠ECD=150°⇒∠ECD=60°
Hence, the correct answer is option (a).