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Consider the problem on khanacademy.

I set up the equation and solve using the Law of Sines as follows.

$$\frac{\sin 40^{\circ}}{30} = \frac{\sin x}{40}$$ $$ \sin^{-1}\left(40\cdot\frac{\sin 40^{\circ}}{30}\right) = x$$ $$x = 58.99^{\circ}$$ From what I understand, arcsin restricts outputs from $-180$ to $180$ degrees, which is why i get $58.99$. My question is what stops $(180-58.99) = 121.01$ from being a valid answer as well.

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  • $\begingroup$ $\arcsin$ range is from $-90^\circ$ to $90^\circ$ $\endgroup$ – kingW3 Mar 11 '17 at 16:54
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The simple answer (in this problem and many others like it) is that you do not know. The "solution" presented in the video is incomplete.

Given the known information (your friend is $40$ feet away, the distance from your friend to the kite is $30$ feet, and the angle between your line of sight to your friend and the line of sight to the kite is $40$ degrees), and accepting the unspoken assumption that all of these things are in one vertical plane (that is, the kite is directly in front of you when you face your friend, not off to the left or right), there are two places where the kite might possibly be, namely, the two intersections of the line from your position at $40$ degrees above with the circle of radius $30$ feet around your friend. At one of those intersections, the angle between the string and your line of sight is $$\arcsin\left(\frac{40}{30}\sin(40^\circ)\right) \approx 58.987^\circ,$$ and at the other the angle is $$180 - \arcsin\left(\frac{40}{30}\sin(40^\circ)\right) \approx 121.013^\circ.$$

In the first case the string makes an angle approximately $81.013^\circ$ with the ground, and in the other the string makes an angle approximately $18.987^\circ$ with the ground. Mathematically, either one of these angles is a possible solution of the given problem, so instead of a unique solution you have a solution set containing two possible angles.

You could apply additional real-world information, such as noticing that the wind is relatively light and having experience that tells you your friend cannot fly a kite in such a wind at an angle of less than $20^\circ$ above horizontal, and therefore rule out the second answer, but you cannot do that using only the information given in the original problem. Moreover, if the numbers in the problem were different in such a way that both of the geometric solutions put the angle of the string at a plausible angle from the ground for the given wind conditions, you would not be able to identify the angle of the string uniquely even with this additional real-world information.

One case where you could determine mathematically that the angle is acute is if the string is longer than the distance between you and your friend. For example, suppose the string were $45$ feet long. In that case, one of the two intersections of the line at an angle $40^\circ$ above the direction from you to your friend and the circle of radius $45$ around your friend would be below you. Of course in the real world kites cannot fly underground, but even without that knowledge, we can say the solution is mathematically unique, since if the kite were at that second intersection point we would have said you were standing at a $140^\circ$ angle of the triangle made by you, your friend, and the kite, not at a $40^\circ$ angle.

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