how to calculate this ... some algebra rule?

**problady** · March 25th 2011, 02:34 AM

guys if I have something like this:

cos(theta)/x=r can I transpose this fomula as x=r*cos(theta) - is this correct?

btw. I am trying to understand if x-adjacent r-hypothenus x=r*cos(theta) -> I want to come up to this part while taking into consideration that cos(theta)=x/r

please help me

**e^(i*pi)** · March 25th 2011, 03:02 AM

edit: my bad,misread

Taken at face value if $r = \dfrac{\cos \theta }{x} \text{ then } x = \dfrac{\cos \theta}{r}$

I suspect you want to show, using a right angled triangle that $x = r \cos \theta$ .

If this is the case recall that by definition $\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$

**problady** · March 25th 2011, 03:38 AM

could you please show me how to go step-by-step from: $\cos \theta = \dfrac{\text{x}}{\text{r}}$
to $x = r \cos \theta$ -> transpose formulae step is very important for me ... to understand what I look for

**e^(i*pi)** · March 25th 2011, 03:40 AM

Multiply both sides by $r$

$r \cos \theta = \dfrac{x}{r} \cdot r = x$

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