okay,
so i remember learning a way to find the inner angles for a triangle in geometry, from only the side lengths. only problem is, ive forgotten how to do it.
i think i remember that there's somehow a relationship between an angle and the side opposite to it, but i cant be sure.
anyway, if anyone knows what im talking about, would you please help me out and remind me the rule.
it would be very much appreciated.
Anonymous
Favorite Answer
If you have a right triangle and have side lengths, then you can use SOACAHTOA to find the measures of the other angles from the sides given.
sinθ = opp/hyp
cosθ = adj/hyp
tanθ = opp/adj
If you have an oblique triangle (a triangle that does not have a 90 degree angle), then you can use the Law of Cosines.
a^2 = b^2 + c^2 - 2bc*Cos(A)
b^2 = a^2 + c^2 - 2ac*Cos(B)
c^2 = a^2 + b^2 - 2ab*Cos(C)
Where A, B, and C are the angles OPPOSITE to the corresponding letter to the sides.
I hope this helps!
Anonymous
Let's say the three angles are A, B and C. Let's also say that the side opposite of the "A" angle has length 'a', the side opposite of the "B" angle has length 'b', and the third side has length 'c'.
The Law of Sines says:
sin(A) / a = sin(B) / b = sin(C)/c
The Law of Cosines says
a^2 = b^2 + c^2 - 2bc cos(A)
b^2 = a^2 + c^2 - 2ac cos(B)
c^2 = a^2 + b^2 - 2ab cos(C)
Take what you know, and use that to find the sides you don't know.
Mathsorcerer
Given a triangle with sides having lengths a, b, and c...
Law of Cosines
c^2 = a^2 + b^2 - 2ab*cos(k) where "k" is the angle formed between sides a and b. From this, we can rearrange to show
cos^-1[(a^2 + b^2 - c^2)/(2ab)] = k
You can use this to find the angle measures.
Captain Matticus, LandPiratesInc
Law of Cosines:
c^2 = a^2 + b^2 - 2ab * cos(C)
b^2 = a^2 + c^2 - 2ac * cos(B)
a^2 = b^2 + c^2 - 2bc * cos(A)
a, b, c represent lengths of the sides
A, B, C represent the corresponding angles.
Just solve for the angle in each case
arccosine [ (a^2 + b^2 - c^2) / (2ab) ] = C
arccosine [ (a^2 + c^2 - b^2) / (2ac) ] = B
arccosine [ (b^2 + c^2 - a^2) / (2bc) ] = A
Anonymous
you need sin cos and tan there :)
SOH CAH TOA
SOH stand for sin = opposite & hypotenuse(if you have the length of the opposite and hypotenuse then you use sine)
CAH stands for cos = adjacent & hypotenuse (if you have the length of the adjacent and hypotenuse then you use cos)
TOA stands for tan = opposite & adjacent (if you have the length of the opposite and adjacent then you use tan)
hope that helps :) xx