Basic trigonometry question


by MechatronO
Tags: basic, trigonometry
MechatronO
MechatronO is offline
#1
Feb23-14, 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.
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SteamKing
SteamKing is offline
#2
Feb23-14, 11:05 AM
HW Helper
Thanks
P: 5,598
What you call the Law of Cosines is actually the Law of Sines.
MechatronO
MechatronO is offline
#3
Feb26-14, 04:22 PM
P: 25
Typo.

More suggestions?


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