SSC CGL 2022 PYQ — Question 71

The correct answer is option 2 i.e. $\frac{19}{√50}$

Solution :

Given that tan² A + 2tanA - 63 = 0

Let tan A =x

Then x² + 2x -63 = 0

x² + 9x - 7x -63 = 0

x(x + 9) - 7(x + 9) = 0

(x - 7)(x + 9) = 0 

x = 7 or x = -9                        (negative value is not considered)

Therefore, tanA = x = 7

i.e. Perpendicular = 7 ; Base = 1

So (hypotenuse)² = (perpendicular)² + (base)²

      (hypotenuse)² = 7² + 1²'

      (hypotenuse)² = 49 + 1

      (hypotenuse)² = 50

      (hypotenuse) = $\sqrt{50}$

Now , sinA = 7/$\sqrt{50}$  ; cos A = 1/$\sqrt{50}$

Required result is  (2sinA + 5cosA) = (2 × 7/$\sqrt{50}$ ) + (5 × 1/$\sqrt{50}$ )

                                                                        = 19/$\sqrt{50}$