Question

An angle is drawn in standard position passing through the unit circle at (0.643,0.766). The angle in standard position θ has a measure of 50°.
What is cos50°?
cos50°≈ __________

Answer 1

Answer:

$\text{cos}(50^{\crc})=0.643$

Step-by-step explanation:

We have been given that an angle is drawn in standard position passing through the unit circle at (0.643,0.766). The angle in standard position θ has a measure of 50°. We are asked to find $\text{cos}(50^{\crc})$.

We know that on unit circle the x-coordinates represent cos and y-coordinates represent sin.

Therefore, the value of $\text{cos}(50^{\crc})$ would be 0.643 as it represent x-coordinate of our given point.

Answer 2

Hello!

The asnwer is: Cos50° ≈ 0.643

Why?

A unit circle is a cirgle with a radius equal of 1, knowing that,  also know the following:

The angle is drawn passing trough the unit circle at (0.643,0.766) it means that:

$x=0.643\\y=0.766$

So,

Cos50° ≈ 0.643

We can prove that by following the next steps:

- If it's a unit circle,here is a right triangle with hypotenuse of 1,

$1^{2}=x^{2}+y^{2}$

$1^{2}=0.643^{2}+0.766^{2}$

$1^{2}=0.4134 +0.5867$

$1=1.0001=1$

- We can determine the cosine of the angle by the following formula:

$cos(\alpha)=\frac{x}{hypotenuse} \\cos(\alpha)^{-1}=cos(\frac{0.643}{1})^{-1} \\\alpha=49.98$

Therefore,

Cos(α)=49.98°≈50°

Also, if there is a right triangle, according to the Pythagorean Thorem:

$1^{2}=(Cos(\alpha))^{2}+(Sin(\alpha))^{2} \\Cos(50)=\sqrt{1-(Sin(50))^{2}}=0.6427$

Hence,

Cos50° ≈ 0.643

Have a nice day!