Given:
In a right angle triangle θ is an acute angle and [tex]\tan\theta =\dfrac{3}{5}[/tex].
To find:
The value of [tex]\cos \theta[/tex].
Solution:
In a right angle triangle,
[tex]\tan \theta=\dfrac{Perpendicular}{Base}[/tex]
We have,
[tex]\tan\theta =\dfrac{3}{5}[/tex]
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,
[tex]Hypotenuse=\sqrt{Perpendicular^2+base^2}[/tex]
[tex]Hypotenuse=\sqrt{(3x)^2+(5x)^2}[/tex]
[tex]Hypotenuse=\sqrt{9x^2+25x^2}[/tex]
[tex]Hypotenuse=\sqrt{34x^2}[/tex]
[tex]Hypotenuse=x\sqrt{34}[/tex]
In a right angle triangle,
[tex]\cos \theta=\dfrac{Base}{Hypotenuse}[/tex]
[tex]\cos \theta=\dfrac{5x}{x\sqrt{34}}[/tex]
[tex]\cos \theta=\dfrac{5}{\sqrt{34}}[/tex]
Therefore, the value of [tex]\cos \theta[/tex] is [tex]\dfrac{5}{\sqrt{34}}[/tex].
