Trigonometry, Circles, Chord length, Area of Segment problem

• Apr 23rd 2010, 05:30 PM
jibba
Trigonometry, Circles, Chord length, Area of Segment problem
http://i42.tinypic.com/rh78u0.jpg

Ok basically i can figure out this equation from this info which is:

8 x (sin a°/2) = 6 x (sin b°/2)

I got the above equations by using the chord length formula as both circles share a similar chord length.

It is pretty much two simultaneous equations subbed into each other.
not sure how to go about solving this though
All help is appreciated
• Apr 23rd 2010, 05:55 PM
Quote:

Originally Posted by jibba
http://i42.tinypic.com/rh78u0.jpg

Ok basically i can figure out this equation from this info which is:

8 x (sin a°/2) = 6 x (sin b°/2)

I got the above equations by using the chord length formula as both circles share a similar chord length.

It is pretty much two simultaneous equations subbed into each other.
not sure how to go about solving this though
All help is appreciated

Hi jibba,

you still need the 2 angles "a" and "b", but then you only have an equality.
That equation doesn't give you either angle.

There is an easy way to solve this, since the triangle from
one point of intersection of the circles to both centres has sides 3, 4 and 5.
This is right-angled since

$3^2+4^2=5^2$

Then, the angle "a" you mentioned is $sin^{-1}\frac{3}{5}=cos^{-1}\frac{4}{5}=tan^{-1}\frac{3}{4}$

Also "b" is $sin^{-1}\frac{4}{5}$ etc.

Now draw the chord and work with the back to back right-angled triangles.

Now that you have the angles, you can calculate the overlapping area
by subtracting the triangle within the circle from the sector,
doing that for both circles and adding the two results.
Then double that to include the part underneath the line joining the circle centres.