Question

Pre Calculus, Trigonometry Help

Answer 1

Answer:

$\displaystyle cos\theta=\frac{36}{164}=\frac{9}{41}$

$\displaystyle tan\theta=\frac{160}{36}=\frac{40}{9}$

$\displaystyle csc\theta=\frac{164}{160}=\frac{41}{40}$

$\displaystyle sec\theta=\frac{164}{36}=\frac{41}{9}$

$\displaystyle cot\theta=\frac{36}{160}=\frac{9}{40}$

Step-by-step explanation:

Trigonometric ratios in a Right Triangle

Let ABC a right triangle with the right angle (90°) in A. The longest length is called the hypotenuse and is the side opposite to A. The other sides are called legs and are shorter than the hypotenuse.

Some trigonometric relations are defined in a right triangle. Being $\theta$ one of the angles other than the right angle, h the hypotenuse, x the side opposite to $\theta$ and y the side adjacent to $\theta$, then

$\displaystyle sin\theta=\frac{x}{h}$

$\displaystyle cos\theta=\frac{y}{h}$

$\displaystyle tan\theta=\frac{x}{y}$

$\displaystyle csc\theta=\frac{h}{x}$

$\displaystyle sec\theta=\frac{h}{y}$

$\displaystyle cot\theta=\frac{y}{x}$

We are given the values of h=164 and x=160, let's find y

$y=\sqrt{164^2-160^2}=36$

Now we compute the rest of the ratios

$\displaystyle cos\theta=\frac{36}{164}=\frac{9}{41}$

$\displaystyle tan\theta=\frac{160}{36}=\frac{40}{9}$

$\displaystyle csc\theta=\frac{164}{160}=\frac{41}{40}$

$\displaystyle sec\theta=\frac{164}{36}=\frac{41}{9}$

$\displaystyle cot\theta=\frac{36}{160}=\frac{9}{40}$