Solving Basic Transposition Problem: Isolating A and Finding its Value

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WildFlower
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Sin180-A = (32.4/390)



I'm usually OK with transposing but I'm having trouble in this example. I need to isolate A and find its value. Can I simply move the Sin across so it becomes Sin-1 and leave 180-A?



For my final answer I get 175.2 degrees which leads me to think I'm of the mark because I suspect the answer should be under 90 degrees.
 
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I really hate this idea of "moving" things from one side of the equation to another! It avoids thinking about what is really going on!

Your equation is sin(180- A)= 32.4/390, right? (The parentheses are a good idea also- there is a difference between that and sin(180)- A.)

On the left side of the equation the sine function is applied to 180- A. To "undo" that you apply it's inverse function, arcsin or sin-1 to both sides of the equation. Yes, it is 180- A= sin-1(32.4/390)= 4.8 degrees so A= 180- 4.8= 175.2 degrees as you say.

But sine is not "one-to-one" so arcsine is a multi-valued function. It is also true that sin(180- A)= sin(A) so A= 4.8 degrees also satisfies that equation.