Maybe it's me, but this didn't seem easy!?!? I'm getting an answer of sin a = 1; is this correct?
Here's a summary of what I did:
You have the cosine of a difference identity:
$\displaystyle cos (a-b) = cos\, a\: cos\, b + sin\, a\: sin\, b \quad$ (Eq1)
Use the Law of sines to write sin b in terms of sin a:
$\displaystyle sin\, b = \frac{3}{4}sin\, a \quad$ (Eq2)
Use the pythagorean identity to write cosine in terms of sine:
$\displaystyle cos\, a = \sqrt{1-sin^2\, a}$ (Eq3a)
$\displaystyle cos\, b = \sqrt{1-sin^2\, b}$ (Eq3b)
Plug (2), (3a), and (3b) into (1), plus the fact that cos (a-b) = 3/4:
$\displaystyle \frac{3}{4} = \sqrt{1-sin^2\, a}\: \sqrt{1-\left(\frac{3}{4}sin\, a\right)^2} + sin\, a\: \left(\frac{3}{4}sin\, a\right)$ (Eq4)
Solved for sin a and I got sin a = 1.
I'm sure I did something wrong here.
In general , if $\displaystyle |ac| = B ~,~ |bc| = A ~,~ \cos(a-b) = X $
The method to find the value of $\displaystyle \sin(a)$ is as follows :
First determine which of the angles $\displaystyle a,b $ is larger by comparing their opposite sides , let's say $\displaystyle a > b $ .
Let $\displaystyle d $ be the other point on line $\displaystyle ab$ such that $\displaystyle |ac| = |cd| = B $
Then we have
$\displaystyle \angle bcd = a-b $ , by using cosine formula ,
$\displaystyle |bd| = \sqrt{A^2 + B^2 - 2ABX} $ ,
Therefore , $\displaystyle \sin(a) : A = \sin(a-b) : |bd| $
$\displaystyle \sin(a) =\frac{A \sin(a-b)}{ \sqrt{A^2 + B^2 - 2ABX} } $