Originally Posted by

**Archie Meade** hi purplec,

is your sketch like the attachment?

then, as skeeter showed

$\displaystyle Sin(65.8^o)=\frac{opposite\ to\ angle}{hypotenuse}=\frac{R}{R+380}$

multiply both sides by the denominator

$\displaystyle (R+380)Sin(65.8^o)=\frac{R(R+380)}{R+380}=R$

multiply out the left

$\displaystyle RSin(65.8^o)+380Sin(65.8^o)=R$

subtract R from both sides to bring them together

$\displaystyle RSin(65.8^o)-R+380Sin(65.8^o)=R-R=0$

subtract the 380Sin part from both sides to have the R terms alone on the left

$\displaystyle RSin(65.8^o)-R+380Sin(65.8^o)-380Sin(65.8^o)=-380Sin(65.8^o)$

$\displaystyle RSin(65.8^o)-R=-380Sin(65.8^o)$

factorise the R parts

$\displaystyle R[Sin(65.8^o)-1]=-380Sin(65.8^o)$

divide both sides by the multiplier of R

$\displaystyle R=\frac{-380Sin(65.8^0)}{Sin(65.8^o)-1}=380\frac{(-1)Sin(65.8^o)}{(-1)[1-Sin(65.8^o)]}=380\frac{Sin(65.8^o)}{1-Sin(65.8^o)}$