Mechanics. Vectors. Find magnitude and direction of forces

[moenste]

Homework Statement

Find the magnitude and direction of the resultant of the following pair of forces:
60 N at 150 degrees to 20 N

Answer:
43.8 N at 136.7 degrees to 20 N and 13.3 degrees to 60 N

Homework Equations

Cosine and sine rules:
R² = a² + b² - 2 * a * b * cos A
a / sin A = b / sin b = c / sin c

The Attempt at a Solution

R² = 60² + 20² - 2 * 60 * 20 * cos 30 = 1921.54
R = 43.8 N

43.8 / sin 30 = 60 / sin X
sin X = 60 sin 30 / 43.8 = 0.68
X = 43.23 degrees or 180 - 43.23 = 136.8 degrees.

From the graph the angle looks definitely like 43.23 degrees and but in the book the answer is 136.8 degrees. My graph looks like

The CA is 60 N, CD is 20 N. ACD is 150 degrees. CB is R = 43.8. CDB is 30 degrees. BCD according to my answer is 43.2 degrees, but in the book that angle should be 136.8 degrees. Any idea how to get the right answer-angle without a graph? In the book examples they state to look at the graph, and other 3 examples have worked correctly. But in this example doesn't matter how big or small I draw the graph I can't get the BCD angle to look as large as 136.8 degrees. Maybe there is a calculus way to check the right angle out of the 43.2 and 136.8 angles?

P. S. The graph is from the web, but mine looks similar to that one. Here ACD is roughly 135 degrees, not 150 as required.

 

[haruspex]

You may be confusing yourself by choosing AC, the shorter of the two sides, to represent the greater force. Wouldn't it make more sense to swap them over?
Clearly the resultant should have a direction closer to the larger of the two forces. You don't specify which angle X is supposed to represent in the diagram.

 

[moenste]

You may be confusing yourself by choosing AC, the shorter of the two sides, to represent the greater force. Wouldn't it make more sense to swap them over?
Clearly the resultant should have a direction closer to the larger of the two forces. You don't specify which angle X is supposed to represent in the diagram.

Hm, I didn't actually think of the sizes, always used a same size for both forces. I shall try it out on a new graph.

Angle X = 43.2 / 136.8 = BCD.

Update: with the new graph everything works. Thank you.