find with with angles

• Sep 14th 2008, 12:17 PM
pyrosilver
find with with angles
Quote:

Find the total grazing area of a goat. The animal is tied to the notheast corner of a 40' by 40' barn, by an 80' rope. The south side of the barn is extended by a fence. Assume that there is grass everywhere except inside the barn.
^^ This totally stumped on.

Quote:

A 16.0-inch chord is drawn in a circle whose radius is 10.0 inches. What is the angular size of the minor arc of this chord? what is the length of the arc, to the nearest tenth of an inch
No idea how to find this!!

• Sep 14th 2008, 12:23 PM
Jhevon
Quote:

Originally Posted by pyrosilver
^^ This totally stumped on.

No idea how to find this!!

did you draw the diagrams for the problems?

note that in the first problem, you are finding the area for the sector of a circle.

in the second problem, use the formula:

$c = 2r \sin \frac {\theta}2$

where $r$ is the radius, $c$ is the length of the chord, and $\theta$ is the angle subtended by the chord
• Sep 14th 2008, 12:49 PM
pyrosilver
Quote:

Originally Posted by Jhevon
did you draw the diagrams for the problems?

note that in the first problem, you are finding the area for the sector of a circle.

in the second problem, use the formula:

$c = 2r \sin \frac {\theta}2$

where $r$ is the radius, $c$ is the length of the chord, and $\theta$ is the angle subtended by the chord

I don't even have the slighest inkling on how to do it, I have no idea what you just said. Thanks for trying to help me, but do you think you could go into a little more depth/explanation?
• Sep 14th 2008, 12:51 PM
Jhevon
Quote:

Originally Posted by pyrosilver
I don't even have the slighest inkling on how to do it, I have no idea what you just said. Thanks for trying to help me, but do you think you could go into a little more depth/explanation?

the first problem you won't understand until you draw the diagram. did you do that?

the second problem is plug-and-chug. i gave you the formula. there is nothing for you to do but plug in numbers and evaluate

the "angle subtended by the chord" is the minor angle they are talking about

i left out the formula for the arc length: $s = r \theta$. $r$ and $\theta$ are the same as before. $s$ is the arc length
• Sep 14th 2008, 12:56 PM
icemanfan
For the second question you also need the formula for arc length, which is $s = r\theta$, where s is the arc length, r is the radius, and $\theta$ is the arc angle in radians. Edit: Jhevon beat me
• Sep 14th 2008, 01:03 PM
Soroban
Hello, pyrosilver!

As Jhevon suggested, make a sketch!

Quote:

Find the total grazing area of a goat.
The animal is tied to the notheast corner of a 40' by 40' barn by an 80' rope.
The south side of the barn is extended by a fence.
Assume that there is grass everywhere except inside the barn.

Code:

              * * *           *:::::|:::::*         *:::::::|:::::::*       *::::::::|80::::::*       :::::::::|:::::::::       *:::::::::|::::80:::*       *----*----*---------*       *::::| 40 |:::::::::*         *::|    |::::::::*     - - - -*- - * - - * - - - -

The upper region is a semicircle with radius 80 feet.

In the lower-left, there is a quarter-circle with radius 40 feet.

In the lower-right, there is a strange-shaped piece.
Code:

    A * - - - - - - - - * D       |  *    80      *   40 |      *       |      30° *    *     B * - - - - - - - * C             40√3

Right triangle $ABC$ has side 40 and hypotenuse 80.

From elementary Trig, $\angle ACB = 30^o = \angle DAC$
. . and hence: $BC = 40\sqrt{3}$

$\Delta ABC$ has base $40\sqrt{3}$ and height $40.$

Sector $ADC$ has $\frac{1}{12}$ of the area of a circle with radius 80.

I assume you can find those separate areas ... and add them.

• Sep 14th 2008, 01:27 PM
pyrosilver
OK I got it now, ty!
• Sep 14th 2008, 02:45 PM
Soroban
Hello again, pyrosilver!

Quote:

A 16-inch chord is drawn is a circle with a radius of 10 inches.

(a) What is the angular size of the minor arc of the chord?

Code:

              * * *           *    |  8  *       A* - - - C - - - *B       *  \    |    /  *             \  |  /10       *      \ |θ/      *       *        *        *       *        O        *       *                *         *              *           *          *               * * *

In right triangle $BCO\!:\;\;OB = 10,\;BC = 8$

Let $\theta = \angle COB$

Then: . $\sin\theta \:=\:\frac{8}{10} \quad\Rightarrow\quad \theta \:=\:\sin^{-1}(0.8) \:=\:0.927295218$ radians

Therefore: . $\angle AOB \;=\;2\theta \;\approx\;1.8546$ radians

Quote:

(b) What is the length of arc (nearest tenth)?

$\text{arc}\,AB \;=\;r\theta \;=\;10(1.8546) \;\approx\;18.5$ inches.