Expert: Paul Klarreich - 1/14/2010

Question
How to calculate the area of triangle when Triangle ABC, a=6 b=3 and Cos(A-B)=4/5.


Answer
Questioner: Ashwin
Country: India
Category: Advanced Math
Private: No
Subject: Mathematics
Question: How to calculate the area of triangle when Triangle ABC, a=6 b=3 and Cos(A-B)=4/5.
............................................

Use these facts:

1. The area of Tr. ABC = 1/2 a b sin C

2. C = 180 - (A + B)

3. cos(A - B) = cos A cos B + sin A sin B

Now we need those trig values.

Law of sines:

 a        b
------ = -------
sin A    sin B

 6        3
------ = -------
sin A    sin B

sin A = 2 sin B
...................
We need all four values.  

cos A = sq(1 - sin^2 A)

cos B = sq(1 - sin^2 B)

cos A = sq(1 - 4 sin^2 B)

cos A cos B = sq(1 - 4 sin^2 B) sq(1 - sin^2 B)

And cos A cos B + sin A sin B = 4/5, so

sq(1 - 4 sin^2 B) sq(1 - sin^2 B) + 2 sin^2 B = 4/5

>>>>>>>>> abbreviate  sin b = s,  do some algebra.

sq(1 - 4 s^2) sq(1 - s^2) + 2 s^2 = 4/5

sq(1 - 4 s^2) sq(1 - s^2)  = 4/5 - 2 s^2

(1 - 4 s^2)(1 - s^2)  = 16/25 + 4 s^4 - 16/5 s^2

1 + 4 s^4 - 5s^2  = 16/25 + 4 s^4 - 16/5 s^2

1  - 5s^2  = 16/25  - 16/5 s^2

25 - 125s^2 = 16 - 80s^2

9 = 45 s^2

1 = 5 s^2

s = 1/sqrt(5)

sin B = sqrt(5),  so cos B = 2/sqrt(5)

Now you should be able to finish up.