Trigonometry angle - calculation failed, can't explain.

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The discussion focuses on solving for angles in a triangle using the Law of Sines and Law of Cosines. The initial calculation for angle A resulted in an incorrect value due to misunderstanding the potential for two solutions in the Law of Sines. It was clarified that the calculated angle A was actually its supplement, and the correct angle can be found by subtracting from 180 degrees. The importance of redrawing the triangle to scale for better visualization was emphasized. The correct method for future calculations involves recognizing the conditions under which multiple solutions may arise.
Monocerotis
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Homework Statement


I have to find the measures of the angles x & y.

2mwtkc6.jpg


Homework Equations


Sine Law
Cosine Law


The Attempt at a Solution




First thing I tried to do was find the measure of the angle @ A.

a/SinA = c/SinC
66/SinA = 25/Sin10.5

and then I end up with 28 degrees for the angle @ A. 28 degrees is obviously wrong. I'm ending up with an error and I can't understand why because so far as I understand my procedure is correct.

I want to solve for angle A, then because I already have angle C I can solve for angle B.

From angle B I could find the angle next to (x), thereby finding x becasue the sum of the two would be 180.
 
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Your picture is way out of scale and is misleading you. Angle A is actually much larger, angle C much smaller, and line AB much shorter than in your picture. Remember the law of sines can give two solutions and this is one of those cases. Your calculation for angle A is actually giving you the supplement of angle A. Subtract it from 180 to get angle A. Redraw your picture more to scale and you will see.
 
LCKurtz said:
Remember the law of sines can give two solutions and this is one of those cases. Your calculation for angle A is actually giving you the supplement of angle A. Subtract it from 180 to get angle A. Redraw your picture more to scale and you will see.

Thanks man, I didn't know that law of sines can give you two solutions, we just started our trig unit last class.

So just to be sure, for future assignments or whatever, I would work out the question like this.

Sin C/c = Sin B/b

Sin 10.5/25 = Sin B/66

(66)Sin10.5/25 = Sin B
Sin^-1(0.4811) = B
28.75 = B

180 - 28.75 = B
151.25 = B

and then work out the rest of the triangle, which is much easier

is that the correct method ?
 
Given triangle with angles ABC with sides a,b,c opposite, suppose you know angle A, side b and side a. If you use the law of sines to find angle B, there will be two solutions whenever a is between b and b sin(a): b > a > b sin(A).

If b is outside that range there will be only one solution. You don't always take the supplement.
 

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