
#1
Feb2314, 08:00 AM

P: 25

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. 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. 



#2
Feb2314, 11:05 AM

HW Helper
Thanks
P: 5,554

What you call the Law of Cosines is actually the Law of Sines.




#3
Feb2614, 04:22 PM

P: 25

Typo.
More suggestions? 


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