In a right-angled triangle ABC, the bisector CD is drawn from the vertex of the right angle.

In a right-angled triangle ABC, the bisector CD is drawn from the vertex of the right angle. Find the angle ADC if the angle B = 32 degrees.

Since CD is the bisector of the right angle C, then ∠ АCD = ∠ DCB = ∠ ACB / 2 = 90 ° / 2 = 45 °.
In any triangle, the sum of the interior angles is 180 °.
Using this fact, for Δ ABC we get: ∠ ABC + ∠ BAC + ∠ ACB = 180 °.
Since ∠ ABC = 32 ° and ∠ ACB = 90 °, then ∠ BAC = 180 ° – ∠ ACB – ∠ ABC = 180 ° – 90 ° – 32 ° = 58 °.
For Δ АDС we have: ∠ АDC + ∠ DАC + ∠ АCD = 180 °, whence ∠ АDC = 180 ° – ∠ АCD – ∠ DАC = 180 ° – 45 ° – 58 ° = 77 °.
Answer: 77 °.