how do I establish the angle of any given corner in a triangle?

Question

I have the length of two sides 3770mm and 1230mm of a triangle how do I establish each of the angles and the length of the other side

Answer 1

you need another bit of information .. is it a right triangle?

even then 3770 could be hypotenuse and other side is
sqrt(3770^2 - 1230^2)
or it could not and the hypotenuse would be
sqrt(3770^2 + 1230^2 )
the angles follow from normal sin = opp / hyp

Answer 2

You have to use the law of cosines, which states:

c^2 = a^2 + b^2 - 2abcosC
b^2 = a^2 + c^2 - 2bccosB
a^2 = b^2 + c^2 - 2accosA

To find c^2, you use the first part of that equation, which is known as the pythagorean theorem.

c^2 = 3770^2 + 1230^2

3966mm is the length of the third side.

15729156 = 14212900 + 1512900 - 2(3770)(1230)cos(C)

15729156 = 15725800 - 9274200cos(C)

3356 = -9274200cos(C)

-.000362 = cos(C)

arccos(-.000362) = C

arccos(.000362) = C

C is approximately pi/2 radians or 90 degrees

Answer 3

That isn't enough information.

The two sides could be at any angle except pi (i.e. in a straight line) and the other side would join the open ends.

Is the type of triangle given? or one of the angles?

Answer 4

that depends, if it's a right triangle, you can use the pythagorean theorem. a^2+b^2=c^2
if the 3770mm is the hypotenuse then your answer is
3770^2-1230^2=(about)3563.7059
if 3770 is not the hypotenuse then
3770^2+1230^2=(about)3965.5769

Answer 5

Thou canst. You need to know, as a minimum, the 3 items required to show congruency.

Answer 6

I'd say use both your sine rule and cosine rule and see if you can form simultaneous equations of some sort.