Question

If $sin\theta =\frac{a}{\sqrt{a^2+b^2}},0<\theta <{90}^{\circ }$, find the values of $cos\theta$ and $tan\theta$.

Answer 1

We have,
$sin\theta =\frac{a}{\sqrt{a^2+b^2}}$

$\therefore cos\theta =\sqrt{1-si{n}^2\theta }=\sqrt{1-\frac{a^2}{a^2+b^2}}=\sqrt{\frac{b^2}{a^2+b^2}}=\frac{b}{\sqrt{a^2+b^2}}$

$\therefore tan\theta =\frac{sin\theta }{cos\theta }\Rightarrow tan\theta =\frac{a/\sqrt{a^2+b^2}}{b/\sqrt{a^2+b^2}}=\frac{a}{b}$