Given:
Right triangle XYZ has right angle Z.
[tex]\sin(x)=\dfrac{12}{13}[/tex]
To find:
The value of [tex]\cos x[/tex].
Solution:
We know that,
[tex]\sin^2(x)+\cos^2(x)=1[/tex]
[tex]\cos^2(x)=1-\sin^2(x)[/tex]
[tex]\cos(x)=\pm\sqrt{1-\sin^2x}[/tex]
For a triangle, all trigonometric ratios are positive. So,
[tex]\cos(x)=\sqrt{1-\sin^2x}[/tex]
It is given that [tex]\sin(x)=\dfrac{12}{13}[/tex]. After substituting this value in the above equation, we get
[tex]\cos(x)=\sqrt{1-(\dfrac{12}{13})^2}[/tex]
[tex]\cos(x)=\sqrt{1-\dfrac{144}{169}}[/tex]
[tex]\cos(x)=\sqrt{\dfrac{169-144}{169}}[/tex]
[tex]\cos(x)=\sqrt{\dfrac{25}{169}}[/tex]
On further simplification, we get
[tex]\cos(x)=\dfrac{\sqrt{25}}{\sqrt{169}}[/tex]
[tex]\cos(x)=\dfrac{5}{13}[/tex]
Therefore, the required value is [tex]\cos(x)=\dfrac{5}{13}[/tex].
