In $\triangle A B C$, $\mathrm{D}$ is a point on $\mathrm{BC}$ such that $\angle B A D=\frac{1}{2} \angle A D C$ and $\angle B A C=77^{\circ}$ and $\angle C=45^{\circ}$. What is the measure of $\angle A D B$ ?
Answer & explanation
Correct answer: option 2
In \(\Delta \)ABC,
\(\angle\)BAC + \(\angle\)ABC + \(\angle\)ACB = 180 [Angle sum property]
= 77 + \(\angle\)ABC + 45 = 180
= 122 + \(\angle\)ABC = 180
= \(\angle\)ABC = (180 - 122)
= \(\angle\)ABC = 58
In \(\Delta \)BAD,
= \(\angle\)ABD + \(\angle\)BAD + \(\angle\)ADB = 180 [Angle sum property]
= 58 + 58 + \(\angle\)ADB = 180
= 116 + \(\angle\)ADB = 180
= \(\angle\)ADB = 180 - 116
= \(\angle\)ADB = \({64}^\circ\)
Therefore, \(\angle\)ADB is \({64}^\circ\).