Let AB and CD be two parallel lines and PQ be a transversal such that PQ intersects AB at the point R and CD at the point S, respectively. If ∠BRP = (2x + 13)° and ∠DSP = (3x − 22)°, then find ∠CSP.

Question

Question

This question was previously asked in

RRB NTPC Graduate Level CBT-I Official Paper (Held On: 06 Jun, 2025 Shift 2)

Answer 1

Given:

AB || CD

PQ is a transversal intersecting AB at R and CD at S.

∠BRP = (2x + 13)°

∠DSP = (3x − 22)°

Formula used:

When two parallel lines are intersected by a transversal:

1. Alternate exterior angles are equal.

2. Angles on a straight line (linear pair) sum up to 180°.

Calculations:

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∠BRP and ∠DSP are alternate exterior angles.

Since AB || CD, we have:

∠BRP = ∠DSP

⇒ 2x + 13 = 3x - 22

⇒ 13 + 22 = 3x - 2x

⇒ 35 = x

Now, substitute the value of x into ∠DSP:

∠DSP = (3 × 35 - 22)°

⇒ ∠DSP = (105 - 22)°

⇒ ∠DSP = 83°

∠CSP and ∠DSP form a linear pair on the line CD.

∠CSP + ∠DSP = 180°

⇒ ∠CSP + 83° = 180°

⇒ ∠CSP = 180° - 83°

⇒ ∠CSP = 97°

∴ ∠CSP = 97°.

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