Question

in the diagram below, \\( \overline { a d } \\) bisects \\( \angle c a b \\), \\( m \angle a d b = 98 ^ { \circ } \\) and \\( m \angle b = 45 ^ { \circ } \\). find \\( m \angle c a b \\).

Explanation:

Step1: Find \(m\angle BAD\)

In \(\triangle ABD\), use the angle - sum property of a triangle (\(m\angle A + m\angle B+m\angle C = 180^{\circ}\)).
Let \(m\angle BAD=x\). We know \(m\angle ADB = 98^{\circ}\) and \(m\angle B = 45^{\circ}\).
So, \(x+45^{\circ}+98^{\circ}=180^{\circ}\).

$$x=180^{\circ}-(45^{\circ} + 98^{\circ})$$

$$x=180^{\circ}-143^{\circ}$$

$$x = 37^{\circ}$$

Step2: Find \(m\angle CAB\)

Since \(AD\) bisects \(\angle CAB\), let \(m\angle CAB = y\). Then \(y = 2m\angle BAD\) (definition of angle - bisector).
Since \(m\angle BAD=37^{\circ}\), then \(y = 2\times37^{\circ}\)

Answer:

\(m\angle CAB = 74^{\circ}\)