Finding areas of sectors using radians

**pturo1** · February 15th 2013, 03:20 AM

There is a question on the area of a sector of a circle on which I am stuck and was wondering whether anyone could give me a pointer.

I have attached an image of the diagram to which the question refers...

The diagram shows an arc of ABC of a circle centre o and radius 10cm.
Angle AOB = $\theta$ radians and angle AOC is a right angle.

a. Write down, in terms of theta, the area of sector AOB.

I can do this easily enough: $area = \frac{1}{2} r^2 * \theta$

area = $50 \theta$

b. Show that the area of the shaded segment is $25(\pi -2\theta - 2cos \theta)cm^2$

This is where I get stuck.

I can find the area of the sector OBC using the formula and get $50(\frac{\pi }{2} - \theta)$ = $ 25\pi - 50 \theta$

But I can't see how to get the area of the triangle.

Having said that, multiplying out $25(\pi - 2\theta - 2 cos \theta)$ gives:

$25\pi - 50\theta -50 cos\theta$ , so the area of the triangle must be $50 cos \theta$ .

I just can't see how to get at this area for the triangle, though. Any advice would be greatly appreciated!

**Plato** · February 15th 2013, 03:27 AM

Originally Posted by pturo1

The diagram shows an arc of ABC of a circle centre o and radius 10cm. Angle AOB = $\theta$ radians and angle AOC is a right angle.
a. Write down, in terms of theta, the area of sector AOB.

I can do this easily enough: $area = \frac{1}{2} r^2 * \theta$
area = $50 \theta$

b. Show that the area of the shaded segment is $25(\pi -2\theta - 2cos \theta)cm^2$

You can use this webpage.

**HallsofIvy** · February 15th 2013, 06:52 AM

That's a isosceles triangle having two sides of length 10 and one angle of measure $\theta$ . If you drop a perpendicular, you divide it into two right triangles having hypotenuse 10 and angle $\theta/2$ . So the two legs of the right triangle have length $10cos(\theta)$ and $10sin(\theta)$ . Since the legs are the "base" and "altitude" of a right triangle, the area of each is $(1/2)(10 cos(\theta/2))(10 sin(\theta/2)$ and the area of the entire triangle is $100 sin(\theta/2)cos(\theta/2)$ .

**pturo1** · February 15th 2013, 11:53 AM

Thank you for your replies. Plato, thanks for the link; it looks interesting but involves cotangents which I have not yet got to in my mathematical education.

Thanks for your explanation Halls of Ivy. I can follow your working but there is just one thing I don't understand (I am probably failing to see something obvious):

How do you know one of the angles in the isocoles triangle is $\theta$ ? The only conclusion I could draw was that the angle would be $\frac{\pi}{2} - \theta$ radians.

Thanks again for your help.

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