Question

for the following point, let $\theta$ be the angle in standard position whose terminal side contains the point. compute $cos(\theta)$ and $sin(\theta)$. $q(9,40)$ $cos(\theta)=$ $sin(\theta)=$

Explanation:

Step1: Calculate the radius $r$

For a point $(x,y)=(9,40)$ in the coordinate - plane, use the formula $r = \sqrt{x^{2}+y^{2}}$. So, $r=\sqrt{9^{2}+40^{2}}=\sqrt{81 + 1600}=\sqrt{1681}=41$.

Step2: Calculate $\cos(\theta)$

The formula for $\cos(\theta)$ is $\cos(\theta)=\frac{x}{r}$. Substituting $x = 9$ and $r = 41$, we get $\cos(\theta)=\frac{9}{41}$.

Step3: Calculate $\sin(\theta)$

The formula for $\sin(\theta)$ is $\sin(\theta)=\frac{y}{r}$. Substituting $y = 40$ and $r = 41$, we get $\sin(\theta)=\frac{40}{41}$.

Answer:

$\cos(\theta)=\frac{9}{41}$, $\sin(\theta)=\frac{40}{41}$