Question

Evaluate cotθ if sinθ= √6/5

Answer 1

Answer: $cot \theta =\sqrt{ \frac{19}{6} }$.

Step-by-step explanation: Given sinθ= √6/5.

We know radio of $Sin\theta =\frac{Opposite \ side }{Hypotenuse}$.

Therefore,

$\frac{Opposite \ side }{Hypotenuse}= \frac{\sqrt{6} }{5}$.

Let us apply Pythagoras Theorem to find the third side(Adjacent side) of the triangle .

$(a)^2 + (b)^2 = (c)^2.$

$(\sqrt{6})^2+(b)^2=(5)^2$

$6+(b)^2 = 25$

Subtracting 6 from both sides, we get

$6-6+(b)^2 = 25-6$

$(b)^2 = 19$

$b=\sqrt{19}$

Therefore, adjacent side = $\sqrt{19}$

We know,

$cot \theta = \frac{Adjacent \ side}{Opposite \ side}$

$cot \theta =\frac{\sqrt{19} }{\sqrt{6} }$

Or

$cot \theta =\sqrt{ \frac{19}{6} }$.