How to Find the Area of a Triangle Using Basic Trigonometry?

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To find the area of a triangle formed by a line at an angle A1 from a centerpoint of a circle with radius R, and at a distance D from the center, the Law of Sines is applied. The user calculates angles A2 and A4 using trigonometric identities, ultimately expressing the area as B*H/2, where H is the height derived from the sine of angle A1. The discussion highlights the complexity of the solution and questions if a simpler method exists. The user acknowledges a typo in referring to the Law of Cosines instead of the Law of Sines. Further suggestions for a more straightforward approach are welcomed.
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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.
 
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What you call the Law of Cosines is actually the Law of Sines.
 
Typo.

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