1. Trig Problem

Hi All,
for a triangle ABC, angle B is 50 degrees. Side AB is twice that of AC. Find angle A.

This means that side AB= x and AC= 2x. Then I thought the only way I could find angle A is if i could find x and then use the Sine rule to find A. But I don't have side AB? Can I assume this is 3x...?

Using the Cosine rule equation: (2x)^2 = (x)^2 + (3x)^2 - 2(x)(3x) Cos50
I continue with this to solve for x...?

Thanks

2. Re: Trig Problem

I think you must have a typo in your problem. If AB=x and AC=2x, then from the law of sines you have:

$\frac { \sin (C)} {2x} = \frac { \sin (B)} {x}$

$\sin(C) = 2 \sin(50) = 1.532$

which is impossible. I wonder if the problem statement should have said: "Side AB is twice that of BC?" If so, then let length BC = x, AB=2x, and AC = y, and from the law of sines you have:

$\frac { \sin (A)} {x} = \frac { \sin (50)} {y}$

and from law of cosines:

$y = \sqrt {x^2 + (2x)^2 - 4x^2 \cos (50) } = x \sqrt { 5-4 \cos 50}$

Sub this back into the previous equation and solve for $\sin A$