SSC CGL 2023 PYQ — Question 71

The correct answer is option 1 i.e. $\frac{\sqrt{3}}{2}$.

We are given that triangle XYZ is right-angled at Y, and sin⁡X = 1/2​.

Let the hypotenuse XZ = 2k.

The side opposite to angle X, XY = k

Using the Pythagorean theorem in triangle XYZ, we find YZ:

YZ = $\sqrt{XZ^2-XY^2} = \sqrt{(2k​)^2-k​^2}= \sqrt{3}k​$

sinZ = cosX = $\frac{\sqrt{3}}{2}$

cosZ = sinX = $\frac{1}{2}$

cos X cos Z + sin X sin Z = $\frac{\sqrt{3}}{2}$ × $\frac{1}{2}$ + $\frac{1}{2}$ × $\frac{\sqrt{3}}{2}$ = $\frac{\sqrt{3}}{4}$+ $\frac{\sqrt{3}}{4}$ = $\frac{2\sqrt{3}}{4}$ = $\frac{\sqrt{3}}{2}$