# Thread: finding an angle in a circle given a point of (0.6, Y)

1. ## finding an angle in a circle given a point of (0.6, Y)

Basic a unit circles is given
______
| / (0.6,Y)
| /
| /
|A)________

Find angle A and Y, two decimal place

So far I got this

Since point is =(cosA,sinA), Y must equal to (Sin angle A)

2. Originally Posted by jepal
Basic a unit circles is given
______
| / (0.6,Y)
| /
| /
|A)________

Find angle A and Y, two decimal place

So far I got this

Since point is =(cosA,sinA), Y must equal to (Sin angle A)
Hi jepal,

$\cos A =.6$

This means that $\cos A=\frac{x}{r}=\frac{6}{10}=\frac{3}{5}$

$x=3$
$r=5$

Use the Pythagorean Theorem to find y.

$r^2=x^2+y^2$

$5^2=3^2+y^2$

$y=4$

So $\sin A = \frac{y}{r}=\frac{4}{5}=.8$

Your point is $(.6, .8)$

Your angle is $\cos^{-1}\frac{3}{5} \ \ or \ \ \sin^{-1}\frac{4}{5}$ or approximately $53.1^{\circ}$

3. Originally Posted by masters
Hi jepal,

$\cos A =.6$

This means that $\cos A=\frac{x}{r}=\frac{6}{10}=\frac{3}{5}$

$x=3$
$r=5$

Use the Pythagorean Theorem to find y.

$r^2=x^2+y^2$

$5^2=3^2+y^2$

$y=4$

So $\sin A = \frac{y}{r}=\frac{4}{5}=.8$

Your point is $(.6, .8)$

Your angle is $\cos^{-1}\frac{3}{5} \ \ or \ \ \sin^{-1}\frac{4}{5}$ or approximately $53.1^{\circ}$
Thank you, but how did u get 10, is it an assumption that radius is 10?

4. Originally Posted by jepal
Thank you, but did I just assume radius is 10, when no radius is given!
What is the fractional form of the decimal number "0.6"?

5. i figured that was the answer, but what is r then? im so confused! wouldnt it be

cos A=x/r

cos A=06./r? no? Please explain. Thanks