1. ## equatios system

Solve the equatios system:

$\displaystyle \left\{ \begin{array}{l} 7\cos \alpha + 9\cos \beta = 10 \\ 7\sin \alpha - 9\sin \beta = 0 \\ \end{array} \right.$

2. Originally Posted by dhiab
Solve the equatios system:

$\displaystyle \left\{ \begin{array}{l} 7\cos \alpha + 9\cos \beta = 10 \\ 7\sin \alpha - 9\sin \beta = 0 \\ \end{array} \right.$
Note that you have

$\displaystyle \left\{ \begin{array}{l} 7 \sin \alpha \cos \alpha + 9 \sin \alpha \cos \beta = 10 \sin \alpha \\ 7\sin \alpha \cos \alpha - 9\sin \beta \cos \alpha = 0 \\ \end{array} \right.$

(Do you see what I did?) Now subtract the second equation from the first. can you finish from there?

3. Add two equations to the system:

$\displaystyle \left\{\begin{array}{l}7\cos\alpha+9\cos\beta=10\\ 7\sin\alpha-9\sin\beta=0\\\sin^2\alpha+\cos^2\alpha=1\\\sin^2\ beta+\cos^2\beta=1\end{array}\right.$

$\displaystyle \sin\beta=\frac{7\sin\alpha}{9}, \ \cos\beta=\frac{10-7\cos\alpha}{9}$

Replace them in the last equation.

$\displaystyle \frac{49\sin^2\alpha}{81}+\frac{100-140\cos\alpha+49\cos^2\alpha}{81}=1\Rightarrow \cos\alpha=\frac{17}{35}\Rightarrow\cos\beta=\frac {77}{105}$

Now find $\displaystyle \sin\alpha, \ \sin\beta$ from the last two equations and choose the sign such as the second equation is satisfied. Then find $\displaystyle \alpha, \ \beta$.

4. Solution by Mathematica

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