Area of a sector without knowing the angle (can't use a calculator)

[Paulo Serrano]

Homework Statement

http://img10.imageshack.us/img10/9651/mathhelp.jpg
RED was given, BLUE was what I was able to find.

What is the area of the shaded part?

Homework Equations

Area of shaded part = area of two triangles + area of sector

The Attempt at a Solution

I was able to find the area of the two triangles easily enough, but I can't think of a way to find the the value of angle AOB without using a calculator. My idea was to find angles AOP and BOQ and subtract them from 180, but again, no calculators allowed.

 

[mgb_phys]

Take a look at the formula for a chord segment.
You don't need a calculator to find sin(x) if you know the lengths of the sides

 

[Paulo Serrano]

(only sane looking formula I found)

Is that it? To clarify, I'm trying to find the angle in order to find the sector area. Are you saying that there is a way to find the area without knowing the angle?

I haven't studied this in quite some years.

 

[slider142]

The problem implicitly gives you the lengths of all three sides of a triangle. Can you think of a formula that will give you the angles within a triangle given all three sides?

 

[mgb_phys]

Draw a line from A-B, there is a formula for the curve+chord shape which involves sin(theta) You can get sin theta from the sides of the triangle

ps. I don't know if this is the answer - it's just something that occurs

 

[Paulo Serrano]

Once again, I fail to notice the simplest things. Triangle APO is a standard 30-60-90 triangle. Triangle BQO an isosceles right triangle, which means it's other two angles are 45 degrees.

With that information I got this:
http://img44.imageshack.us/img44/9651/mathhelp.jpg

Knowing the angle, I can find my answer. Thanks a lot to both of you for pushing me in the right direction.

----

EDIT: I know the answer now but I have a question about how to write it that one of you may be able to help me with.

Area of Shaded Part = http://www4b.wolframalpha.com/Calculate/MSP/MSP361966i5a8g78bbbh200003b56bi9282i8195d?MSPStoreType=image/gif&s=10 + http://www4b.wolframalpha.com/Calculate/MSP/MSP541966i5a8g757454900001ih9c1ah3gdch55g?MSPStoreType=image/gif&s=10 + [PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP1921966hg7d1iggfhie000042hif6h986h19e32?MSPStoreType=image/gif&s=21

I'm studying for an exam in which calculators are not allowed. Is there a way to reduce this further? I can only get to:

http://www4b.wolframalpha.com/Calculate/MSP/MSP10901966e00a08553fg700004ed479a898634aad?MSPStoreType=image/gif&s=30 + [PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP1921966hg7d1iggfhie000042hif6h986h19e32?MSPStoreType=image/gif&s=21 + 1

 

[slider142]

You have there a multiple of a transcendental number added to a multiple of an irrational added to an integer. It's not much of a simplification to complete the addition by finding a common denominator, ie., it's fine the way it is.