Question

In the adjoining figure, $AD=BD$ $\angle ABD={65}^o$ & $\angle DAC={22}^o$.
Find $\angle ACD$

Answer 1

$AD=BD$

$\angle BAD=\angle ABD=65$

$\angle BAC=\angle BAD+\angle DAC$

$\angle BAC=65+22$

$\angle BAC=87$

$\angle A+\angle B+\angle C=180°$

$87+65+\angle C=180°$

$\angle C=180-152$

$\angle C=28$

$\therefore \angle ACD=28$