# Question #62169

Jun 4, 2016

$2 \frac{2}{15}$

#### Explanation:

Given $\sin \theta = \frac{3}{5}$
we know that in a right triangle as given in the figure below

$S \in \theta = \text{opposite"/"hypotenuse} = \frac{3}{5}$
From Pythagoras theorem
$\text{adjacent } = \sqrt{{5}^{2} - {3}^{2}}$
$\text{adjacent } = \sqrt{25 - 9} = \sqrt{16} = 4$

From figure
$\cos \theta + \cot \theta = \text{adjacent "/"hypotenuse"+"adjacent "/"opposite}$
$= \frac{4}{5} + \frac{4}{3}$
$= \frac{4 \times 3 + 4 \times 5}{15}$
$= \frac{32}{15}$
$= 2 \frac{2}{15}$