Question

$\text{In the given figure, AD divides}\angle BAC\text{in the ratio 1:3 and AD=DB.Determine the value of x .}$

Answer 1

ANSWER:
∠BAC+∠CAE=180° ∵BE is a straight line

⇒∠BAC+108°=180° ⇒∠BAC=72°
Now, divide 72° in the ratio 1 : 3.
∴a+3a=72°⇒a=18°∴a=18° and 3a=54°
Hence, the angles are ${18}^{0}and{54}^0$
∴∠BAD=18° and ∠DAC=54°
Given,
AD=DB

⇒∠DAB=∠DBA=18°
In ∆ABC, we have:
∠BAC+∠ABC+∠ACB=180° (Sum of the angles of a triangle)

⇒72°+18°+x°=180°⇒x°=90°

∴ x=90