- #1
skrat
- 748
- 8
Homework Statement
I ran across this equation: $$ T_1\cos \vartheta =-T_2\cos \alpha \sin \varphi + T_2\sin \alpha \cos \varphi$$
Get ##\alpha## if ##T_1=80##, ##T_2=50##, ##\vartheta =60^°## and ##\varphi =30^°##.
Homework Equations
The Attempt at a Solution
There are two possible ways:
a) One could simply write $$\sin^2=1-\cos^2\alpha$$ and get a quadratic equation for ##\cos \alpha## with solutions $$(\cos\alpha)_1=0.1196$$ and $$(\cos\alpha)_2=-0.9196.$$ Each solution has two angles ##\alpha##. First solution is $$\alpha_1=\pm 83.13^°$$ and second $$\alpha _2=\pm 156.87^°$$
b) One could use some trigonometric identities and rewrite the equation as $$T_1 \cos\vartheta = T_2\sin(\alpha-\varphi)$$ which means (according to my understanding...) $$\alpha =\varphi+arcsin(\frac{T_1}{T_2}\cos\vartheta)+n\pi.$$ For ##n=0## the solution is (as above in case a)) ##\alpha_1= 83.13^°##, however the problem here is that I can't figure out how to get the other solution in this case.This is absolutely the most embarrassing question when you realize you are close to a second Bachelor degree...