Question

If sin theta =root 2 by root 3 and lies in the first quadrant,the value of tan theta is

Answer 1

Answer:

$\sqrt{2}$

Explanation:

Let us assume a right triangle ABC, right angled at B such that BC is perpendicular and AB is the base with AC as Hypotenuse

Now,

sin(θ) =  $\frac{\sqrt{2} }{\sqrt{3}}$

therefore let us assume perpendicular BC = $\sqrt{2}x$

then Hypotenuse AC = $\sqrt{3}x$

Now base = $\sqrt{AC^{2} - BC^{2}}$

                 = $\sqrt{(\sqrt{3}x)^{2} - (\sqrt{2}x)^{2}}$

                 = $\sqrt{3x^{2} - 2x^{2} }$ = $\sqrt{1x^{2} } = 1x$

Therefore Base AB = $1x$

Now,

tan(θ) = $\frac{BC}{AB}$

          = $\frac{\sqrt{2}x}{1x} = \frac{\sqrt{2} }{1} = \sqrt{2}$

Therefore,

tan(θ) = $\sqrt{2}$

Answer 2

Answer:

root 2 is the answer

use Pythagoras theorem