In Vector chapter i found the formula, $$ \tan\theta=\cfrac{Q\sin\alpha}{P+Q\cos\alpha} $$
Suppose I have the values of $$ Q = |\vec{Q}|\\ P = |\vec{P}|\\ \theta=angle\ between\ \vec{P}\ and\ (\vec{P}+\vec{Q}) $$
How can I find the value for $ \alpha $ ?
So far i have tried this $$ \begin{align} \cfrac{P}{Q}\tan\theta = \sin\alpha-\cos\alpha\tan\theta \end{align} $$

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    $\begingroup$ Here is Wolfram's solution, but isn't nice looking. $\endgroup$ – Hakim Jun 9 '14 at 17:21
  • $\begingroup$ Thanks. But isn't there any solution simpler as i know $ \alpha \ge 0^\circ\ and\ \alpha \le 180^\circ $ $\endgroup$ – palatok Jun 9 '14 at 17:31

From your last relation $$ \frac{P}{Q}\sin\theta=\sin\alpha\cos\theta-\cos\alpha\sin\theta=\sin(\alpha-\theta) $$ and supposing $\left|(P/Q)\sin\theta\right|\leq1$ $$ \alpha=\theta+\arcsin\left(\frac{P}{Q}\sin\theta\right)+2k\pi\\ \alpha=\theta+\pi-\arcsin\left(\frac{P}{Q}\sin\theta\right)+2k\pi $$


P, Q and P+Q denote the magnitude of three vectors $\vec{P}^{}$,$\vec{Q}^{}$ and $\vec{P}^{}+\vec{Q}^{}$ as mentioned.
$\theta$ be the angle between two vectors $\vec{P}^{}$ and $\vec{P}^{}+\vec{Q}^{}$.
$ \alpha $ be the angle between the vector sets $\vec{P}^{}$ and $\vec{Q}^{}$.
$ \gamma $ be the angle between the vector sets $\vec{P}^{}$ and $\vec{P}^{}+\vec{Q}^{}$ as shown in the following figure.

lazy dog

From the properities of a triangle, $$\beta = \pi - \alpha$$ (relationship between exterior and interior angles of a vertex in a triangle).
Hence, by Lami's theorem, the value of $\alpha$ can be determined by the following formula,$$ \frac{P+Q}{sin(\beta)}=\frac{P}{sin(\gamma)}=\frac{Q}{sin(\theta)}$$ and from the previous equation, this equation reduces in following way.$$ \frac{P+Q}{sin(\pi - \alpha)}=\frac{Q}{sin(\theta)}$$ $$ sin(\alpha)= sin(\theta)\left(1+\frac{P}{Q}\right)$$ Answer: $$ \alpha= arcsin\left(sin(\theta)\left(1+\frac{P}{Q}\right)\right)$$


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