1. ## Trigonometric system

Hi,

I can't solve this trigonometric system in order to y and only appearing Theta(1), h, L and w:

sin(Theta(1))= x/L1

sin(Theta(2))= (w-x)/L2

cos(Theta(1))= y/L1

cos(Theta(2))=(y-h)/L2

L1+L2=L

If necessary you can use sin(theta(1))=sin(theta(2))=w/L

Thanks

2. Hello, kolg!

I can't solve this trigonometric system in order to y
and only appearing Theta(1), h, L and w. .
What does that mean?

. . $\displaystyle [1]\;\;\sin(\theta_1) \:=\: \frac{x}{L_1}$

. . $\displaystyle [2]\;\;\sin(\theta_2) \:=\: \frac{w-x}{L_2}$

. . $\displaystyle [3]\;\;\cos(\theta_1) \:=\: \frac{y}{L_1}$

. . $\displaystyle [4]\;\;\cos(\theta_2) \:=\: \frac{y-h}{L_2}$

. . $\displaystyle [5]\;\;L_1+L_2\:=\:L$

If necessary, you can use: $\displaystyle \sin(\theta_1)\,=\,\sin(\theta_2)\,=\,\frac{w}{L}$

This doesn't make sense.
It may be a case of TMI (too much information).

I will assume that we are to solve for $\displaystyle y.$

Square [1]: .$\displaystyle \frac{x^2}{L_1^2} \:=\:\sin^2(\theta_1)$

Square [3]: .$\displaystyle \frac{y^2}{L_1^2} \:=\:\cos^2(\theta_1)$

Add: . $\displaystyle \frac{x^2}{L_1^2} + \frac{y^2}{L_1^2} \;=\;\underbrace{\sin^2(\theta_1) + \cos^2(\theta_1)}_{\text{This is 1}}$

We have: .$\displaystyle \frac{x^2+y^2}{L_1^2} \:=\:1 \quad\Rightarrow\quad w^2 + y^2 \:=\:L_1^2 \quad\Rightarrow\quad y^2 \:=\:L_1^2 - x^2$

Therefore: .$\displaystyle y \;=\;\pm\sqrt{L_1^2 - x^2}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Oh-oh!

I think they want the answer in terms of $\displaystyle L$ . . . and not $\displaystyle L_1$

3. Thanks for the reply, but yes I need that in order to L, and that is what I can't figure out

Thanks