SSC CHSL 2021 PYQ — Question 63

The correct answer is option 1 i.e. 7

Given:

  $\angle$PQR = 135°

PQ = 8√2 cm and

PR = 17 cm

Concept USed:

$\cos\theta = \frac{b^2 + c^2 - a^2}{2bc}$

Explanation:

Let the length of QR be x cm.

According to the given question,

$\cos135^0 = \frac{x^2 + (8\sqrt{2})^2 - (17)^2}{2x(8\sqrt{2})}$

⇒ $-{1 \over \sqrt{2}} = \frac{x^2 + 128 - 289}{16\sqrt{2}x}$

⇒ $-16x = x^2 - 161$

⇒ x² + 16x - 161 = 0

⇒ x² + 23x - 7 x - 161 = 0

⇒ x(x + 23) - 7(x + 23) = 0

⇒ (x + 23)(x - 7) = 0

⇒ x + 23 = 0 ; x - 7 = 0

⇒ x = -23 (not possible); x = 7

Hence the required value of QR is x = 7 cm.