Given a right-angled triangle ABK with a right angle B. Points C and D lie on sides AB and AK, respectively

Given a right-angled triangle ABK with a right angle B. Points C and D lie on sides AB and AK, respectively, CD is parallel to BK, point P lies on AD. What is the ACP angle if the PCD angle is 60 °?

Given:

angle (B) = 90,

Point (C) belongs to side AB,

Point (D) belongs to the AK side,

CD ll BK,

Point (P) belongs to the AD side.

Since CD ll BK, we get an identical right-angled triangle, only smaller. The resulting triangle (ACD) is similar to the originally given triangle (ABC). From this it follows that angle (C) = angle B = 90.

angle (PCD) = 60,

angle (C) = angle (PCD) + angle (ACP),

90 = 60 + angle (ACP),

angle (ACP) = 90 – 60 = 30,

Answer: angle (ACP) = 30.