Given:
The point is (-6,5).
It is on the terminal side of theta.
To find:
The values of [tex]\sin\theta[/tex] and [tex]\cos \theta[/tex].
Solution:
In the point (-6,5),
x-coordinate : x=-6
y-coordinate : y=5
So, [tex]r=\sqrt{x^2+y^2}[/tex]
[tex]r=\sqrt{(-6)^2+(5)^2}[/tex]
[tex]r=\sqrt{36+25}[/tex]
[tex]r=\sqrt{61}[/tex]
Now,
[tex]\sin \theta=\dfrac{y}{r}[/tex]
[tex]\sin \theta=\dfrac{5}{\sqrt{61}}[/tex]
[tex]\cos \theta=\dfrac{x}{r}[/tex]
[tex]\cos \theta=\dfrac{-6}{\sqrt{61}}[/tex]
Therefore, the required values are [tex]\sin \theta=\dfrac{5}{\sqrt{61}}[/tex] and [tex]\cos \theta=-\dfrac{6}{\sqrt{61}}[/tex].
