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SSC CGL 2025 PYQ — Question 5

SSC CGL Tier II - 20 January 2025 - Shift I • Quantitative Aptitude • Q5

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Let PQR be a right angled triangle, right-angled at R. Let PQ = 29 cm, QR = 21 cm and ∠Q = θ. Find the value of cos 2 θ – sin 2 θ.

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Correct answer: D

Explanation

The correct answer is option 4 i.e. 41841\frac{41}{841}.

PQ = 29 cm, QR = 21 cm and ∠Q = θ

PQR is a right-angled triangle, right-angled at R.

PR = √(PQ2 - QR2) = √(292 - 212) = √(841 - 441) = √400 = 20 cm

cos θ = Base/Hypotenuse = QR/PQ = 21/29

sin θ = Perpendicular/Hypotenuse = PR/PQ = 20/29

cos2 θ – sin2 θ = (21/29)2 - (20/29)2 = (441/841) - (400/841) = 41/841

Therefore, the required value is 41/841.