# Area of a Sector in Radians.

• Sep 2nd 2009, 08:55 PM
evinyssen
Area of a Sector in Radians.
Hi, I need some help with the following two problems:

1.) Find an exact value for the fraction of the sector represented by the triangle AOB in the sector AOB in the diagram.

http://img216.imageshack.us/img216/2715/21950781.jpg

2). The area of the sector AOB and the triangle AOB are at a ratio of 3:2.

The angle AOB is in radians.

Show that 2θ-3sinθ=0.

http://img197.imageshack.us/img197/4268/70433372.jpg

I have managed to get:

3=½r²θ and 2=½r²sinθ

Therefore:

½r²θ-3=0 and ½r²sinθ-2=0

But I'm unsure where to go from there.

Any help would be appreciated. Thanks very much.
• Sep 2nd 2009, 10:14 PM
pacman
For #1.

Area triangle AOB = ([(3^1/2)/4]r^2)

Area sector AOB = (1/2)(r^2)(Angle in radians)

Area triangle AOB/ Area sector AOB = ([(3^1/2)/4]r^2)/(1/2)(r^2)(pi/3),

r^2 will cancel and 2 and simplifying

= 3[(3)^1/2]/2pi
• Sep 3rd 2009, 05:33 AM
Hello evinyssen

Welcome to Math Help Forum!
Quote:

Originally Posted by evinyssen
2). The area of the sector AOB and the triangle AOB are at a ratio of 3:2.

The angle AOB is in radians.

Show that 2θ-3sinθ=0.

http://img197.imageshack.us/img197/4268/70433372.jpg

I have managed to get:

3=½r²θ and 2=½r²sinθ See my comment below.

Therefore:

½r²θ-3=0 and ½r²sinθ-2=0

But I'm unsure where to go from there.

Any help would be appreciated. Thanks very much.

Thanks for showing us your working. You're almost there!

You're right in saying that the area of the sector is $\tfrac12r^2\theta$, and that the area of the triangle is $\tfrac12r^2\sin\theta$. But you shouldn't then say that they are equal to $3$ and $2$ respectively. They are in the ratio
$3:2$,which is a different thing.

We can always write a ratio as a fraction, so do just that:

$\frac{\tfrac12r^2\theta}{\tfrac12r^2\sin\theta}=\f rac{3}{2}$

Cancel the fraction on the LHS:

$\frac{\theta}{\sin\theta}=\frac{3}{2}$

$\Rightarrow 2\theta = 3\sin\theta$