# Math Help - finding length in truss

1. ## finding length in truss

Find the lengths of BF and CF?

i have found that BF is 4779 i now need to know how to find CF

2. Hello, Zak!

Find the lengths of BF and CF.

i have found that BF is 4779 ... i now need to know how to find CF.

We need repeated applications of the Law of Sines.

In triangle $ABF:\;\frac{\sin(\angle BFA)}{3000} \,=\,\frac{\sin52^o}{4779}$

. . $\sin(\angle BFA) \:=\:\frac{3000\sin52^o}{4779} \:=\:0.494670906$

. . $\angle BFA \:=\:29.64805186\:\approx\:29.6^o$

Then: . $\angle BFC\:=\:90^o - 29.6^o\:=\:60.4^o$

In triangle $BCF:\;\frac{\sin(\angle BCF)}{4779} \:=\:\frac{\sin60.4^o}{4500}\$

. . $\sin(\angle BCF) \:=\:\frac{4779\sin60.4^o}{4500} \:=\:0.923403615$

. . $\angle BCF \:=\:67.42885176^o\:\approx\:67.4^o$

Then: . $\angle CBF\:=\:180^o - 60.4^o - 67.4^o \:=\:52.2^o$

Finally: . $\frac{CF}{\sin52.2^o} \:=\:\frac{4500}{\sin60.4^o}\quad\Rightarrow\quad CF \:=\:\frac{4500\sin52.2^o}{\sin60.4^o} \:=\;4089.382738$

Therefore: . $\boxed{CF \:\approx\:4089}$