Question

If $\csc \theta =\frac{13}{5}$, find $\tan \theta$ and $\cos \theta$.

Answer 1

Given:- $\csc \theta =\frac{13}{5}.....\left(1\right)$

As we know that,

$\csc \theta =\frac{\text{hypotenuse}}{\text{Perpendicular}}.....\left(2\right)$

On comparing $\left(1\right)$ and $\left(2\right)$, we get

Hypotenuse $\left(h\right)=13$

Perpendiculatr $\left(p\right)=5$

Base $\left(b\right)=$

Using pythagoras theorem, we have

$h^2=p^2+b^2$

$\Rightarrow b^2={\left(13\right)}^2-{\left(5\right)}^2$

$\Rightarrow b=\sqrt{169-25}=\sqrt{144}=12$

Therefore,

$\tan \theta =\frac{p}{b}=\frac{5}{12}$

$\cos \theta =\frac{b}{h}=\frac{12}{13}$