# Thread: Help with equation of angle x

1. ## Help with equation of angle x

I'm trying to understand how to solv x.

Would be glad if someone could give me a hand.

2. Hello Reno
Originally Posted by Reno
I'm trying to understand how to solv x.

Would be glad if someone could give me a hand.
Look at the diagram I've attached, where I've divided the angles $x$ and $A$ into two parts each.

Here's what you need to find $x$:
$P = \arctan\left(\frac{b}{a}\right)$

$Q = \arctan\left(\frac{a}{b}\right)$

$S = A - Q$

$\frac{c}{\sin S}= \frac{\sqrt{a^2+b^2}}{\sin(180^o- (R+S))}$

$x = P+R$
Can you put all this together?

3. Thanks alot. But I must say that my skills has been better, I would be happy if you could help me further.

4. 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 $x$ in terms of $a, b$ and $A$. It will be pretty complicated if we put it all together. I think it's probably best left in terms of $P,Q,R$ and $S$.

So, as I said in my previous post:
$\frac{c}{\sin S}=\frac{\sqrt{a^2+b^2}}{\sin(180^o-(R+S))}$
(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 $\sin\theta = \sin(180^o-\theta)$:
$\frac{c}{\sin S}=\frac{\sqrt{a^2+b^2}}{\sin(R+S)}$

$\Rightarrow \sin(R+S) = \frac{(\sqrt{a^2+b^2})\sin S}{c}$

$\Rightarrow R = \arcsin\left(\frac{(\sqrt{a^2+b^2})\sin S}{c}\right)-S$
Then I should leave the answer as:
$x = P+R$, where
$P = \arctan\left(\frac{b}{a}\right)$

$R = \arcsin\left(\frac{(\sqrt{a^2+b^2})\sin S}{c}\right)-S$

$S = A-Q$

$Q = \arctan\left(\frac ab\right)$ (or you could simply say $Q = 90^o-P$.)
But you could combine these into a single expression by substituting for each of the variables $P,Q,R,S$ if you wish.

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