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I was doing some math work, and have to solve two trig equations simultaneously, but have no idea how to approach this, can anyone help, just need to be pointed in the right direction.

I have to solve for $T_1$ and $T_2$ using these equations:

$$T_1\sin(45.0)\;+\;T_2\sin(30.0)\;=\;0.00$$ $$T_1\cos(45.0)\;+\;T_2\cos(30.0)\;=\;1720$$

Can someone just tell me the first one or two steps and help me get started, I can't find this in my textbook anywhere. I tried rearranging the equations, didn't really help much.

Edit: Solved, just subbed values for sine and cosine!

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Do you know a value you can substitute in for the sine and cosine terms? – Michael Dyrud Oct 25 '12 at 20:26
Hmm, oh, that can work! Thanks, i'll edit that then! – Link Oct 25 '12 at 20:27
No problem! In general, whenever a trig problem has nice angles involved, i.e. 30,60,90,45, see if putting in the actual values can help at all. – Michael Dyrud Oct 25 '12 at 20:29
wow, it did work got it! thank you. – Link Oct 25 '12 at 20:38
up vote 5 down vote accepted

This is not a trig problem. Even if the angles were ugly, as long as there is no variable under the $\sin$, it's just an obscure constant. If these trigonometric constants bother you, you are free to substitute them with symbols of your choice, as long as you remember to put back the real values in the final result.

Let $a = \sin(45), b = \sin(30), c = \cos(45), d = \cos(30)$ $$\begin{cases}aT_1 + bT_2 = 0 \\ cT_1 + dT_2 = 1720\end{cases} \implies T_1 = \frac{1720b}{bc-ad},~ T2 = \frac{-1720a}{bc-ad}$$ Finally, substitute back the $a, b, c, d$ to get $T_1 = -1720 \frac{\sqrt{2}(1+\sqrt{3})}{2}$ and $T_2 = 1720(1+\sqrt{3})$.
Hopefully I didn't get it wrong...

While it may look like a silly way to solve something, I wrote it to illustrate that the method of solving this problem does not rely on whether the constants involve trigonometric functions or whether their real values are known in any way.

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