# How do you sketch the angle whose terminal side in standard position passes through (1,sqrt3) and how do you find sin and cos?

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#### Explanation

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#### Explanation:

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Feb 9, 2018

$\sin \left(x\right) = \frac{y}{H y p} = \frac{\sqrt{3}}{2}$ and $\cos \left(x\right) = \frac{x}{H y p} = \frac{1}{2}$

#### Explanation:

If you place the triangle on a graph, with one side laying on the positive side of the x-axis, and set the other side to pass through the origin and the given point $\left(1 , \sqrt{3}\right)$ it will form a right triangle.

The bottom side will be the "adjacent" side to the angle, while the vertical side is the "opposite."

knowing the acronym SOH CAH TOA
$\sin = \frac{O p p o s i t e}{H y p o t e \nu s e}$ , $\cos = \frac{A \mathrm{dj} a c e n t}{H y p o t e \nu s e}$ , $\tan = \frac{O p p o s i t e}{A \mathrm{dj} a c e n t}$

and the Pythagorean Theorem
${\left(H y p o t e \nu s e\right)}^{2} = {\left(O p p o s i t e\right)}^{2} + {\left(A \mathrm{dj} a c e n t\right)}^{2}$

We can find that
$H y {p}^{2} = {\left(\sqrt{3}\right)}^{2} + {1}^{2}$
By solving for Hyp, we see that the hypotenuse is 2 units long.

Knowing this, we can then solve the sine and cosine.

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