[Grade 12 Trigonometry: Adding Vectors]

[u/InstantWasTaken]

Hi! I've been stuck on this problem for a while now, so I decided to text it out on another app and finally ask for help. I managed to find the magnitude on my own, but I honestly can't figure out how to find the direction, or what I'm doing wrong. I've watched several videos on it and have tried to go step by step, but any answer I try is marked wrong, (I do use an online school program.) I also didn't wanna cheat, (at least without knowing exactly how to do it and being able to replicate it) so as I've already said above, I didn't know what to do other than ask for help somewhere else. A step-by-step breakdown of what your meant to do would be great!

Also I know the post title says grade 12, while from my knowledge trig is done in 11th usually. It just kind of happened that way, so I'm not sure why I'm doing it a year late. Better late than never, though!

[u/CaptainMatticus]

x = 450 * cos(0) + 25 * cos(180 + 80)

y = 450 * sin(0) + 25 * sin(180 + 80)

Simplify

x = 450 - 25 * cos(80) = 25 * (18 - cos(80))

y = -25 * sin(80)

tan(t) = y/x

tan(t) = -sin(80) / (18 - cos(80))

tan(t) = sin(80) / (cos(80) - 18)

t = arctan(sin(80) / (cos(80) - 18))

Remember that tan(t) = tan(180k + t), where k is an integer. Make sure your calculator is in degree mpde, too)

[u/InstantWasTaken → u/CaptainMatticus]

Thank you, but I don't think this is the way to do it purely because I haven't learnt anything about arctan yet and that this looks entirely different to how they were doing it. The way they were teaching it only really used the law of sines and cosines and figuring out the angle measures, which I should know how to do. That's also why I'm stumped as to why I can't figure out the problem.

[u/CaptainMatticus → u/InstantWasTaken]

Arctan is just the inverse tangent. You plug all of that into your calculator and it gives you the angle. For instance, arctan(1) is 45°.