[RESOLVED] Trigonometry... and bad too!

[BillGeek]

Hey Guys

Another question that, well, is basically Trigonometry based. (I think... )

Suppose you have, as the following illustration:
BAC = 25deg.
AC = 1meter
and A = (2, 1)



How would one calculate the coordinates of point C on the sketch above?

As I never completed any tertiary matematical courses, (and have forgotten my highschool maths already) I have figured out a way to do it that's... well... not right.

Theta = 25deg.
25 / 90 = 0.27
65 / 90 = 0.72

Therefore,
y = y + 0.72
x = x - 0.27

I know, I know, it sucks. That's not right and I know it.

Please guys, I hope that this will be the last question I have to post today.

[zaza]

You already know the y coordinate of C, because you have been given the length AB and the coordinates of A. To figure out the length BC you need to know that the tan of an angle is Opposite / Adjacent. You know the angle and the Adjacent. Then you can figure out the x coordinate of C, because you again know the x distance from A and the start coordinate.

[krtxmrtz]

Coordinates of B: (2,2)

Coordinates of C: y is the same as that of B, i.e. y_c = 2

As for x,

x_c = x_a - CB

But tan(25) = CB / BA so:

CB = BA*tan(25) = (y_b - y_a) tan(25)

And finally,

x_c = x_a - (y_b - y_a) tan(25) = 2 - (2 - 1)*tan(25) = 2 - tan(25) = 1.5337 approx.

[BillGeek]

You already know the y coordinate of C, because you have been given the length AB and the coordinates of A.

WAIT!!! Sorry guys, a bit of a typo in the example above! It's not AB that's 1 meter, it's AC... Sorry...

[opus]

OK, that easy:
You have AC=1 and the angle CAB=25° . In that case CB is sin(25°)=0.422618
[since AC=1 the Sin is equal to the Length !]
Knowing that CAB=25° and ABC=90°, therefor BCA=65° and that makes BA sin(65°)=0.90631
You can check the values by using the a^2 +b^2=c^2!!

I hope you can do the calculation of the coordinates on your own ;-)

[BillGeek]

Ahhh... I keep forgetting which is adjacent and which is opposite, from which angle to work from, which function to use...

Thanks opus. I believe that this will help me a great deal!

Theoretically speaking, if the angle of ACB was 40deg, that would mean that to calculate BA, I would take BA = Sin(40) = 0.745113160479349

Thanks again!

[zaza]

Ahhh... I keep forgetting which is adjacent and which is opposite, from which angle to work from, which function to use...

Thanks opus. I believe that this will help me a great deal!

Theoretically speaking, if the angle of ACB was 40deg, that would mean that to calculate BA, I would take BA = Sin(40) = 0.745113160479349

Thanks again!

If you have a right-angled triangle, then the Hypotenuse is the longest side opposite the right angle, the Opposite side is the side opposite the angle you know / are looking for and the Adjacent side is the one which is adjacent to aforementioned angle, i.e. the one which, with the hypotenuse, actually forms the angle.

sin = Opp / Hyp
cos = Adj / Hyp
tan = Opp / Adj



zaza