Question

If

Answer 1

Given:

In a right angle triangle θ is an acute angle and $\tan\theta =\dfrac{3}{5}$.

To find:

The value of $\cos \theta$.

Solution:

In a right angle triangle,

$\tan \theta=\dfrac{Perpendicular}{Base}$

We have,

$\tan\theta =\dfrac{3}{5}$

It means the ratio of perpendicular to base is 3:5. Let 3x be the perpendicular and 5x be the base.

By using Pythagoras theorem,

$Hypotenuse=\sqrt{Perpendicular^2+base^2}$

$Hypotenuse=\sqrt{(3x)^2+(5x)^2}$

$Hypotenuse=\sqrt{9x^2+25x^2}$

$Hypotenuse=\sqrt{34x^2}$

$Hypotenuse=x\sqrt{34}$

In a right angle triangle,

$\cos \theta=\dfrac{Base}{Hypotenuse}$

$\cos \theta=\dfrac{5x}{x\sqrt{34}}$

$\cos \theta=\dfrac{5}{\sqrt{34}}$

Therefore, the value of $\cos \theta$ is $\dfrac{5}{\sqrt{34}}$.