In triangle ABC the angle at C is 14 degrees and AD is the bisector of angle A, with angle BAD equal to 28 degrees. Find angle ADB in degrees.

Step-by-step solution

1. 1

   Use the bisector to recover the full angle A. A bisector splits $\angle A$ into two equal parts, and one of them is given:

   $\angle BAD = \angle DAC = 28^{\circ} \quad\Longrightarrow\quad \angle A = 2 \times 28^{\circ} = 56^{\circ}$

   This is the whole point of the given $28^{\circ}$: it is a half-angle, not the angle itself.

2. 2

   Apply the angle sum in triangle ABC.

   $\angle A + \angle B + \angle C = 180^{\circ} \quad\Longrightarrow\quad 56^{\circ} + \angle B + 14^{\circ} = 180^{\circ}$

   $\angle B = 110^{\circ}$

   The triangle is obtuse at $B$, which is consistent with the two given angles being small.

3. 3

   Move to the sub-triangle ABD. The point $D$ lies on side $BC$, so the angle at $B$ inside triangle $ABD$ is the same $\angle ABC = 110^{\circ}$ — the ray $BA$ and the ray $BC$ (which contains $BD$) are unchanged.

4. 4

   Close the angle sum in triangle ABD.

   $\angle ADB = 180^{\circ} - \angle BAD - \angle ABD = 180^{\circ} - 28^{\circ} - 110^{\circ} = 42^{\circ}$

5. 5

   Cross-check via the other sub-triangle. In triangle $ACD$: $\angle DAC = 28^{\circ}$ and $\angle C = 14^{\circ}$, so $\angle ADC = 180^{\circ} - 42^{\circ} = 138^{\circ}$. Since $\angle ADB$ and $\angle ADC$ are supplementary along the straight line $BC$, $\angle ADB = 180^{\circ} - 138^{\circ} = 42^{\circ}$ — the same answer by an independent route. (Equivalently, the exterior-angle theorem gives $\angle ADB = \angle DAC + \angle C = 28^{\circ} + 14^{\circ} = 42^{\circ}$ directly.)