Answer
$\sin\theta$ = $ \frac{6}{\sqrt 85}$
$\cos\theta$ =$ \frac{-7}{\sqrt 85}$
$\tan\theta$ = - $ \frac{6}{7}$
Work Step by Step
Given $\theta$ is in standard position. Finding a Point P on terminal side of $\theta$, in the given diagram-
We find point P (-7, 6)
Now, we may apply Definition I to find required trigonometric functions-
We got $ x = -7, y = 6$
Therefore r= $\sqrt (x^{2} + y^{2})$
= $\sqrt ((-7)^{2} + 6^{2})$
= $\sqrt (49 + 36)$
= $\sqrt 85$
i.e. $ x = -7, y = 6,$ and $ r= \sqrt 85$
Applying Definition I-
$\sin\theta$ =$ \frac{y}{r}$ = $ \frac{6}{\sqrt 85}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{-7}{\sqrt 85}$
$\tan\theta$ =$ \frac{y}{x}$ =$ \frac{6}{-7}$ = - $ \frac{6}{7}$
