How Do You Solve for g in the Equation a = 1/(1+C) g sin θ?

[einsteinette]

1. The problem statement.


a = 1/(1+C) g sine pheta


Rearrange equation to solve for g.


My attempt of the solution is:


g = (a+c)/sine pheta

What confuses me is whether or not g sine pheta are all clumped together or if it's just g x sine pheta. Thanks!

 

[rock.freak667]

Well gsinθ = g*sinθ

although I can't tell if your question is


$$a = \frac{1}{(1+C)g sin \theta} \ or \ a=\frac{1}{1+C}* gsin \theta$$

 

[einsteinette]

Ah, ok, thanks. It's the second one.

 

[drizzle]

Ah, ok, thanks. It's the second one.

if so, then:

g = (a+c)/sine pheta

that's not correct,

$$\ a=\frac{1}{1+C}* gsin \theta$$

$$\ a{(1+C)}=gsin \theta$$
$$\ {(a+aC)}= gsin \theta$$
then
$$\ g=\frac{a+aC}{sin \theta}$$