You are Here: Home

# Working out angles without calculator (complex numbers)

Announcements Posted on
1. Hi guys,

If I have a triangle, of which I know all the sides (I'm working with complex numbers) im trying to work out the ARGUMENT i.e arctan (y/x) without a calculator.
This is just the angle between my hypotenuse which is 2 root 2 and my adjacent which =2. Opposite is also 2.

How can I do this?

Thanks, really stuck.
2. If you're looking for angles, you want to start with the cosine rule. Square all the sides and look at what that tells you.
3. That's half a right angle.
4. Sounds like you're dealing with a right-angled isosceles triangle (i.e. half a square) - since the opposite and adjacent are both 2, and the triangle obeys Pythagoras' theorem.
Should be obvious what the angle is...
5. Isn't the answer: arg= tan-1 1= (pi)/4 = 45o
6. Yes ok, I gave a simple example, but I want a general method that I can use for everything. For example, the next question has hypotenuse = 2, adjacent = 1, and opposite = root 3

Thanks
7. (Original post by dannylfc_1)
Yes ok, I gave a simple example, but I want a general method that I can use for everything. For example, the next question has hypotenuse = 2, adjacent = 1, and opposite = root 3

Thanks
It's been suggested...the cosine rule. In general for a triangle with integral sides the angles will not be rational
8. (Original post by dannylfc_1)
Yes ok, I gave a simple example, but I want a general method that I can use for everything. For example, the next question has hypotenuse = 2, adjacent = 1, and opposite = root 3.
These is no general method that will work for everything.

You should know (*) the values of for any integer multiple of or .

All the examples you have given fall into this category, and this is likely no coincidence.

(*) Or be able to work out from a value you do know. e.g. I might not know directly what sin(3pi/4) is, but I know that sin(3pi/4) = sin(pi - 3pi/4) = sin(pi/4) = sqrt(2)/2.
9. (Original post by dannylfc_1)
Yes ok, I gave a simple example, but I want a general method that I can use for everything. For example, the next question has hypotenuse = 2, adjacent = 1, and opposite = root 3

Thanks
okay, but you didn't tell which angle you want to find (ie. angle between hyp and opp or angle between hyp and adj). Yea I also think that you should you cosine rule. Anyway, I don't see what both your questions have to do with complex number...?
10. (Original post by pauching)
okay, but you didn't tell which angle you want to find (ie. angle between hyp and opp or angle between hyp and adj). Yea I also think that you should you cosine rule. Anyway, I don't see what both your questions have to do with complex number...?
yes sorry - the angle between adjacent and hypotenuse.
I see the cosine rule has been mentioned but can someone spell this out for me exactly how you work it out. Maths isn't my strongest subject!
11. (Original post by DFranklin)
These is no general method that will work for everything.

You should know (*) the values of for any integer multiple of or .

All the examples you have given fall into this category, and this is likely no coincidence.

(*) Or be able to work out from a value you do know. e.g. I might not know directly what sin(3pi/4) is, but I know that sin(3pi/4) = sin(pi - 3pi/4) = sin(pi/4) = sqrt(2)/2.
Yes, I understand, so how then do I work out arctan? Thanks
12. If you know the values for sin and cos, you know the value for tan...
13. (Original post by dannylfc_1)
yes sorry - the angle between adjacent and hypotenuse.
I see the cosine rule has been mentioned but can someone spell this out for me exactly how you work it out. Maths isn't my strongest subject!
In a triangle where the side of length is opposite angle A,
14. (Original post by pauching)
okay, but you didn't tell which angle you want to find (ie. angle between hyp and opp or angle between hyp and adj). Yea I also think that you should you cosine rule. Anyway, I don't see what both your questions have to do with complex number...?
I also should have said minus root 3 not root 3

Its because my complex number is z=1-i root 3 and i want to put this in the form re^i(theta). Where r is the hypotenuse which is 2 and theta is the angle arctan. Hope that helps
15. A squared = B squared + C squared - (2BC) x Cos A

just rearrange the formula to make cos A the subject of the equation and you can work out each one
16. Yes thanks, i understand the cosine rule method. Thanks everyone for helping really appreciated
17. Have to say, I strongly disagree with everyone advocating the use of the cosine formula.
18. (Original post by DFranklin)
Have to say, I strongly disagree with everyone advocating the use of the cosine formula.
How else would you do it? I realise I'm getting to the point where I have cos A = x. But then I still need to know how to do the next part in my head, to which I think i might just have to memorise the unit circle.
19. If you have a right angled triangle, you don't need the cosine formula. This is basic (GCSE, I would have thought) trig.
20. True

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: March 23, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### AQA psychology unofficial markscheme

Find out how you did here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams