Arcs/Arc Length

**goliath** · November 16th 2009, 11:30 PM

Hey guys, another question for ya:

0 = theda

Suppose that 0 = 150 degrees

a. Find the length of the circular arc corresponding to 0(theda) on a circle of radius 5.
(For this question I got, 13.09, so could someone check to see if it's correct?)

b. Find the area of the slice corresponding to 0(theda) on a circle of radius 5

c. Find sin 0 and cos 0 (again 0 = theda)

d. Find the angle that the radial line segment of the unit circle ending at (sin 0, cos 0) makes with the positive horizontal axis.

I would GREATLY appreciate it if you guys could also provide the answers just for proofing and rechecking.

**Grandad** · November 17th 2009, 05:19 AM

Hello goliath

Originally Posted by goliath

Hey guys, another question for ya:

0 = theda

Suppose that 0 = 150 degrees

a. Find the length of the circular arc corresponding to 0(theda) on a circle of radius 5.
(For this question I got, 13.09, so could someone check to see if it's correct?)

b. Find the area of the slice corresponding to 0(theda) on a circle of radius 5

c. Find sin 0 and cos 0 (again 0 = theda)

d. Find the angle that the radial line segment of the unit circle ending at (sin 0, cos 0) makes with the positive horizontal axis.

I would GREATLY appreciate it if you guys could also provide the answers just for proofing and rechecking.

If $\theta$ is measured in radians, then the formulae for arc length, $s$ , and area of sector, $A$ , are:

$s =r\theta$

$A = \tfrac12r^2\theta$

In this question $\theta = 150^o=\frac{5\pi}{6}$ radians and $r = 5$ .

So:

(a) Arc length $s = r\theta=5\times \frac{5\pi}{6}\approx 13.09\text{ cm}$ . So you are right!

(b) Area of sector $A = \tfrac12r^2\theta=\frac{5^2\cdot5\pi}{2\cdot6}\app rox32.7\text{ cm}^2$

(c) $\sin\left(\frac{5\pi}{6}\right)=\sin\left(\pi - \frac{5\pi}{6}\right)=\sin\left(\frac{\pi}{6}\righ t)=\frac12$

$\cos\left(\frac{5\pi}{6}\right)=-\cos\left(\pi - \frac{5\pi}{6}\right)=-\cos\left(\frac{\pi}{6}\right)=-\frac{\sqrt3}{2}$

(d) We want the point whose coordinates are $\left(\frac12,-\frac{\sqrt3}{2}\right)$ . Sketch a diagram, then, and you'll see that the point is in the fourth quadrant, with the radial line making an angle $60^o$ below the $x$ -axis. So the angle we want is $-60^o$ (or $300^o$ ), measuring angles positive anticlockwise.

Grandad

**goliath** · November 17th 2009, 04:50 PM

Thank you so much, I appreciate the help!

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