# What is the value of $\alpha$ in, $\tan\theta=\frac{Q\sin\alpha}{P+Q\cos\alpha}.$

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}

• Here is Wolfram's solution, but isn't nice looking. – Hakim Jun 9 '14 at 17:21
• Thanks. But isn't there any solution simpler as i know $\alpha \ge 0^\circ\ and\ \alpha \le 180^\circ$ – 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.
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: $$\alpha= arcsin\left(sin(\theta)\left(1+\frac{P}{Q}\right)\right)$$