# Solving for x: A/sin(x)=B/sin(C-x)

• May 1st 2011, 05:34 AM
ShouldKnowBetter
Solving for x: A/sin(x)=B/sin(C-x)
Hi,

I am trying to solve this equation: A/sin(x)=B/sin(C-x)
for x, where 0<=C<=pi and A<B.

It's been years since I last would have contemplated solving such an expression and quite frankly I am somewhat saddened that I have forgotten how. I once knew how to tackle this. Much of the maths I learnt studying engineering just never got used (Worried).

It actually is a derivative of the law of sines equation relating to solving the torpedo fire control problem.

x= angle of deflection
A= target speed
B= torpedo speed
C= track angle

I ultimately want to be able to use Excel to create and plot curves like these.

Once I am on the right track I will be fine but I can't actually say what approach needs to be taken to deal with this problem. Need someone talking maths to me using maths language again. I can't even remember what this type of problem/expression is called.

Thanks for helping out.

Cheers

SKB

PS: Not sure if this is in the right forum
• May 1st 2011, 06:29 AM
e^(i*pi)
$\displaystyle A\sin(C-x) = B\sin(x)$

$\displaystyle A(\sin(C) \cos(x) - \cos(C) \sin(x)) = B\sin(x)$

$\displaystyle \sin(C) \cot(x) - \cos(C) = \dfrac{B}{A}$

$\displaystyle \cot(x) = \csc(C) \left(\dfrac{B}{A} + \cos(C)\right)$

$\displaystyle \tan(x) = \sin(C) \dfrac{1}{\left(\dfrac{B}{A} + \cos(C)\right)}$

$\displaystyle x = \arctan \left( \dfrac{\sin(C)}{\left(\dfrac{B}{A} + \cos(C)\right)}\right)$
• May 1st 2011, 07:07 AM
ShouldKnowBetter
OK, as I kind of suspected, subbing in trignometric identities was the way to go. Actually, I would have preferred if you just said something like:

Use a basic trig identity substitution (hint: sin(A-B)=sinA.cosB-cosA.sinB)
Relax, it's not complex.

...and let me take it from there and salvage what dignity I have left. (Blush)

Still, thank you!