Hello Reno
First, a rather belated Welcom to Math Help Forum! Originally Posted by
Reno Thanks alot. But I must say that my skills has been better, I would be happy if you could help me further.
I'm assuming that we have to find an expression for $\displaystyle x$ in terms of $\displaystyle a, b$ and $\displaystyle A$. It will be pretty complicated if we put it all together. I think it's probably best left in terms of $\displaystyle P,Q,R$ and $\displaystyle S$.
So, as I said in my previous post:$\displaystyle \frac{c}{\sin S}=\frac{\sqrt{a^2+b^2}}{\sin(180^o-(R+S))}$
(Do you understand how I got this? I used the Sine Rule on the lower triangle, having first used Pythagoras' Theorem on the upper triangle.)
So, noting that $\displaystyle \sin\theta = \sin(180^o-\theta)$:$\displaystyle \frac{c}{\sin S}=\frac{\sqrt{a^2+b^2}}{\sin(R+S)}$
$\displaystyle \Rightarrow \sin(R+S) = \frac{(\sqrt{a^2+b^2})\sin S}{c}$
$\displaystyle \Rightarrow R = \arcsin\left(\frac{(\sqrt{a^2+b^2})\sin S}{c}\right)-S$
Then I should leave the answer as:
$\displaystyle x = P+R$, where
$\displaystyle P = \arctan\left(\frac{b}{a}\right)$
$\displaystyle R = \arcsin\left(\frac{(\sqrt{a^2+b^2})\sin S}{c}\right)-S$
$\displaystyle S = A-Q$
$\displaystyle Q = \arctan\left(\frac ab\right)$ (or you could simply say $\displaystyle Q = 90^o-P$.)
But you could combine these into a single expression by substituting for each of the variables $\displaystyle P,Q,R,S$ if you wish.
Grandad