0 = 3[cos(35)*cos(A) - sin(35)*sin(A)] - cos(A)

[Fresh4Christ]

I have the problem:

0 = 3[cos(35)*cos(A) - sin(35)*sin(A)] - cos(A)
where A is alpha...my unknown degree.

somehow that turns into this:

tan(A) = [cos(35) - 1/3] / sin(35)

I am not drawing the connection or seeing how that is happening...

Could you help? THANKS

 

[robert Ihnot]

Looking at the form we want, the first thing to do is divide by cos(A). Then at that point it is possible to separate tan(A) from the other terms.

 

[theperthvan]

Rearrange to give

$$cos(35).cos(A) - \frac{cos(A)}{3} = sin(35)sin(A)$$

$$cos(A)(cos(35) - \frac{1}{3}) = sin(35).sin(A)$$

divide both sides by $$cos(A)$$ to give

$$(cos(35) - \frac{1}{3}) = sin(35).tan(A)$$

divide both sides by $$sin(35)$$ to give

$$\frac{(cos(35) - \frac{1}{3})}{sin(35)} = tan(A)$$

 

[theperthvan]

oops, that y should be A

EDIT: ignore that

 

[Fresh4Christ]

Thank you soooo much!