- #1

McFluffy

- 37

- 1

**1. Find α(β) given that the sum of the 2 sides= ##(x+y)## and its third, ##z## is a constant for 0<β<180.**

You can imagine that there's two pieces of string connected between two points. One string is as long as the distance between the two points while the other string is longer. If you stretch the longer string such that it will form a triangle like in the diagram, you'll notice that ##x, y## varies but their sum is a constant. The question is how would you find the angle of α given that you know what β, z and the sum of ##x+y## are.

**2. My attempt.**

Before we do it, let’s identify the third angle of the triangle, call it ##n## (in radians)##=\pi-\alpha-\beta=\pi-(\alpha+\beta)##. This will be useful later. We could just apply law of sine straightforwardly but nah. Let’s try and find it instead.

Let’s call the dashed line ##h##. So, ##\sin{\alpha}=\dfrac{h}{y}## and that ##\sin{\beta}=\dfrac{h}{x}##. Solve for ##h## in those two equations so we can get the law of sine. You’ll get the following : ##x\sin{\beta}=y\sin{\alpha}=h##. From this, divide across the equations by ##\sin{\beta}\sin{\alpha}## to get: ##\dfrac{x}{\sin{\alpha}}=\dfrac{y}{\sin{\beta}}=\dfrac{h}{\sin{\alpha}\sin{\beta}}##. You can invert those equations if you want, either way there’s the law of sine. We’d found ourselves a connection between the angle ##\alpha## and ##\beta##, but what about the third angle, ##n##? Let’s try it out, From the law of sine: ##\dfrac{z}{\sin{n}}=\dfrac{z}{\sin{(\pi-(\alpha+\beta))}}##. From symmetry of how the sine function works, that last equation is the same as: ##\dfrac{z}{\sin{(\alpha+\beta)}}##.

So far, we’ve found 4 equations that are equal to each other; a result from the law of sine:##\dfrac{x}{\sin{\alpha}}=\dfrac{y}{\sin{\beta}}=\dfrac{h}{\sin{\beta}\sin{\alpha}}=\dfrac{z}{\sin{(\alpha+\beta)}}##. This is good because we have this: ##\dfrac{h}{\sin{\beta}\sin{\alpha}}=\dfrac{z}{\sin{(\alpha+\beta)}}.## Now all we need to do is find what ##h## is and plug it in. After that, I can now solve for ##\alpha##. If you had figured out what ##x+y## is in terms of trig functions, sine function to be more specific, it’ll be: ##x=\dfrac{h}{\sin{\beta}},y=\dfrac{h}{\sin{\alpha}}##. If you add ##x+y##, it should give you ##x+y=c =\dfrac{h(\sin{\alpha}+\sin{\beta})}{\sin{\alpha}\sin{\beta}}##. We’d define ##x+y=c## so it can be less messy. Solve for ##h##, ##h=c\dfrac{\sin{\alpha}\sin{\beta}}{\sin{\alpha}+\sin{\beta}}##.Plug this into ##\dfrac{h}{\sin{\beta}\sin{\alpha}}=\dfrac{z}{\sin{(\alpha+\beta)}}## and you’ll get:## \dfrac{z}{\sin{(\alpha+\beta})}=\dfrac{c}{\sin{\alpha}+\sin{\beta}}##. We have gathered all the important variables into this one neat equation. Now to solve for ##\alpha## ….

Well this is where I'm stuck.

To find ##\alpha## in ## \dfrac{z}{\sin{(\alpha+\beta})}=\dfrac{c}{\sin{\alpha}+\sin{\beta}}##, we could expand the terms by using angle addition rule for sine and gather all the ##\alpha## terms on the other side:##c\sin{\alpha}\cos{\beta}-z\sin{\alpha}+c\sin{\beta}\cos{\alpha}=z\sin{\beta}##, factor out ##\sin{\alpha}## on the left side to get: ##\sin{\alpha}(c\cos{\beta}-z)+c\sin{\beta}\cos{\alpha}=z\sin{\beta}## And notice that the left hand side of the equation can be represented as the expansion of the angle addition sine rule of ##\alpha+k## where ##k## is well the one we need to figure out too. But I don't know how to proceed now.