- #1
MechatronO
- 30
- 1
We got a circle with a radius R.
From a distance D from the centerpoint a line is inserted at an offset angle A1 from a line drawn though the centerpoint C of the circle, see the picture below.
http://cdn.imghack.se/medium/0861cdab13b8957018f8e167342f2b8e.png
I would like the are of the red triangle, provided D, R and A1.
I drew another triangle with one of its corner in the circles center for help, extracted the angle A2, got H and could then solve the problem but. However I wonder if there is a neater way than this.
What I did:
A3 = 180°-A1
Law of cosines give
sin (A3) / R = (sin A4) /D
A4 = arcsin( sin(A3) * D/R ) = arcsin( sin(180°-A1) * D/R)
A2 = 180° - A3 - A4 = 180° - (180°-A1) - arcsin( sin(180°-A1) * D/R) =
= A1 - arcsin( sin(A1)* D/R)
sin(A1) = H/R
H= R*sin(A1)
cos(A2) = (B+D)/R
B= R*cos(A2) -D
The red area = B*H/2 = (R*cos(A2)-D)*R*sin(A2)/2
And so forth. However, I get the feeling that this solution is more complicated than necessary?
EDIT: Btw the circle hasn't got much to do with the problem, but I'm just using it in a next step.
From a distance D from the centerpoint a line is inserted at an offset angle A1 from a line drawn though the centerpoint C of the circle, see the picture below.
http://cdn.imghack.se/medium/0861cdab13b8957018f8e167342f2b8e.png
I would like the are of the red triangle, provided D, R and A1.
I drew another triangle with one of its corner in the circles center for help, extracted the angle A2, got H and could then solve the problem but. However I wonder if there is a neater way than this.
What I did:
A3 = 180°-A1
Law of cosines give
sin (A3) / R = (sin A4) /D
A4 = arcsin( sin(A3) * D/R ) = arcsin( sin(180°-A1) * D/R)
A2 = 180° - A3 - A4 = 180° - (180°-A1) - arcsin( sin(180°-A1) * D/R) =
= A1 - arcsin( sin(A1)* D/R)
sin(A1) = H/R
H= R*sin(A1)
cos(A2) = (B+D)/R
B= R*cos(A2) -D
The red area = B*H/2 = (R*cos(A2)-D)*R*sin(A2)/2
And so forth. However, I get the feeling that this solution is more complicated than necessary?
EDIT: Btw the circle hasn't got much to do with the problem, but I'm just using it in a next step.