In the given figure, ABC is a non-isosceles right angled triangle, right angled at B. If PQ and BM are parallel and BM = AM then ∠CPQ is equal to?

Correct option is C

Given:

ABC is a non-isosceles right-angled triangle, right angled at B.

PQ and BM are parallel and BM = AM.

Concept Used:

Angles opposite to the equal sides are equal.

Sum of all angles of triangle = 180 degrees

Solution:

Let the angle CPQ be x degrees.

So, Angle CPQ = Angle CBM

Because BM ∥ PQ

As angle B = 90 degrees

So, Angle ABM = (90 – x) degrees

As given in the question,

BM = AM

Therefore, Angle BAM = Angle ABM = (90 – x) degrees

As per the property of triangle,

Angle BAM + Angle PCQ + Angle ABC = 180°

Angle BAM + Angle PCQ + 90°=180°

90 – x + Angle PCQ + 90 = 180

Angle PCQ = x

Hence, Angle PCQ = Angle CPQ = x degrees.