Triangles

An easy way to get $sin(A)=\frac{2}{3}$ is to make $CB=2$ and $AB=3$ .

From Pythagorean theorem, we get $x=\sqrt{3-4}=\sqrt{5}$ .

So the tangent is $tan(B)=\frac{\sqrt{5}}{2}$ .

$\tan^2x + 1 = \sec^2x \\ \dfrac{1}{\cos^2x} = 1 + \tan^2x \\ \dfrac{1}{\cos^2x} = 1 + \tan^2x \\ \dfrac{1}{1 - \sin^2x} = 1 + \dfrac{1}{\tan^2x} \\ \dfrac{1}{1 - \dfrac{4}{9}} -1 = \dfrac{1}{\tan^2x} \\ \dfrac{1}{\dfrac{5}{9}} - 1 = \dfrac{1}{\tan^2x} \\ \dfrac{1}{\tan^2x} = \dfrac{9}{5} - 1 = \dfrac{4}{5} \\ \tan x = \dfrac{\sqrt[]{5}}{2}$