Q:

Evaluate cotθ if sinθ= √6/5

Accepted Solution

A:
Answer: [tex]cot \theta =\sqrt{ \frac{19}{6} }[/tex].
Step-by-step explanation: Given sinθ= √6/5.We know radio of [tex]Sin\theta =\frac{Opposite \ side }{Hypotenuse}[/tex].Therefore, [tex]\frac{Opposite \ side }{Hypotenuse}= \frac{\sqrt{6} }{5}[/tex].Let us apply Pythagoras Theorem to find the third side(Adjacent side) of the triangle .[tex](a)^2 + (b)^2 = (c)^2.[/tex][tex](\sqrt{6})^2+(b)^2=(5)^2[/tex][tex]6+(b)^2 = 25[/tex]Subtracting 6 from both sides, we get[tex]6-6+(b)^2 = 25-6[/tex][tex](b)^2 = 19[/tex][tex]b=\sqrt{19}[/tex]Therefore, adjacent side = [tex]\sqrt{19}[/tex]We know, [tex]cot \theta = \frac{Adjacent \ side}{Opposite \ side}[/tex][tex]cot \theta =\frac{\sqrt{19} }{\sqrt{6} }[/tex]Or [tex]cot \theta =\sqrt{ \frac{19}{6} }[/tex].