SSC CHSL 2023 PYQ — Question 54

The correct answer is option 3 i.e. ${2\sqrt42}\over 11$.

cosθ = Base/Hypotenuse = 11/17

Using Pythagoras theorem, H² = P² + B²

⇒ 17² = P² + 11²

⇒ P² = 289 - 121

⇒ P² = 168

⇒ P = $2\sqrt{42}$

Now, tanθ = Perpendicular/Base = ${2\sqrt42}\over 11$