In a triangle ABC, C = 90˚, and A = 70˚, CD-bisector. Find the angles ∆BCD.

Since the angle is C = 90 °, the ABC triangle is rectangular. It is known that the angle A = 70 °. One of the properties of a triangle says that the sum of all its angles is 180 °. Based on this, we find the angle B:

B = 180 ° – A – C;

B = 180 ° – 70 ° – 90 °;

B = 180 ° – 160 °;

B = 20 °;

The bisector CD divides the angle C into two equal angles:

ACD = DCB = 1/2 C;

ACD = DCB = 1/2 90 °;

ACD = DCB = 45 °;

CDB = 180 ° – B – DCB;

CDB = 180 ° – 20 ° – DCB;

CDB = 180 ° – 20 ° – 45 °;

CDB = 180 ° – 65 °;

CDB = 115 °.

Answer: The angles in the DCB triangle are 20 °, 45 °, 115 °.