trigonometry system of equation

**n0083** · October 13th 2008, 07:42 PM

a, b, and c are known and greater than zero, solve for theta and phi from the two equations i) and ii).
i)a*cos(theta) = b*cos(phi)
ii)a*sin(theta) + b*sin(phi) = c

thanks,

**Soroban** · October 14th 2008, 12:26 PM

Hello, n0083!

The problem is not difficult . . . The answers are ugly!

$a, b, c$ are known and greater than zero. .Solve for $\theta$ and $\phi$

. . $\begin{array}{cccc} a\cos(\theta) &=& b\cos(\phi) & {\color{blue}[1]} \\ a\sin(\theta) &=& c - b\sin(\phi) & {\color{blue}[2]}\end{array}$

$\begin{array}{ccccc}\text{Square {\color{blue}[1]}:} & a^2\cos^2\!\theta &=& b^2\cos^2\!\phi \\<br /> <br /> \text{Square {\color{blue}[2]}:} & a^2\sin^2\!\theta &=& c^2 - 2bc\sin\phi + b^2\sin^2\!\phi \end{array}$

$\text{Add: }\;a^2\underbrace{(\sin^2\!\theta+\cos^2\!\theta)} _{\text{This is 1}} \;=\;c^2 -2bc\sin\phi + b^2\underbrace{(\sin^2\phi + \cos^2\!\phi)}_{\text{This is 1}}$

. . $a^2\;=\;c^2 - 2bc\sin\phi + b^2 \quad\Rightarrow\quad\sin\phi \;=\;\frac{b^2+c^2-a^2}{2bc}$ .[3]

Hence: . $\boxed{\phi \;=\;\sin^{-1}\left(\frac{b^2+c^2-a^2}{2bc}\right)}$

Substitute [3] into [2] and solve for $\sin(\theta)$ . . . then solve for $\theta.$

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