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Finding areas of sectors using radians

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 = 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 =

b. Show that the area of the shaded segment is

This is where I get stuck.

I can find the area of the sector OBC using the formula and get

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

Having said that, multiplying out gives:

, so the area of the triangle must be .

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

Re: Finding areas of sectors using radians

Quote:

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 =

b. Show that the area of the shaded segment is

You can use this webpage.

Re: Finding areas of sectors using radians

That's a isosceles triangle having two sides of length 10 and one angle of measure . If you drop a perpendicular, you divide it into two right triangles having hypotenuse 10 and angle . So the two legs of the right triangle have length and . Since the legs are the "base" and "altitude" of a right triangle, the area of each is and the area of the entire triangle is .

Re: Finding areas of sectors using radians

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 ? The only conclusion I could draw was that the angle would be radians.

Thanks again for your help.