Find the lengths of BF and CF?
i have found that BF is 4779 i now need to know how to find CF
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 $\displaystyle ABF:\;\frac{\sin(\angle BFA)}{3000} \,=\,\frac{\sin52^o}{4779} $
. . $\displaystyle \sin(\angle BFA) \:=\:\frac{3000\sin52^o}{4779} \:=\:0.494670906$
. . $\displaystyle \angle BFA \:=\:29.64805186\:\approx\:29.6^o$
Then: .$\displaystyle \angle BFC\:=\:90^o - 29.6^o\:=\:60.4^o$
In triangle $\displaystyle BCF:\;\frac{\sin(\angle BCF)}{4779} \:=\:\frac{\sin60.4^o}{4500}\$
. . $\displaystyle \sin(\angle BCF) \:=\:\frac{4779\sin60.4^o}{4500} \:=\:0.923403615$
. . $\displaystyle \angle BCF \:=\:67.42885176^o\:\approx\:67.4^o$
Then: .$\displaystyle \angle CBF\:=\:180^o - 60.4^o - 67.4^o \:=\:52.2^o$
Finally: .$\displaystyle \frac{CF}{\sin52.2^o} \:=\:\frac{4500}{\sin60.4^o}\quad\Rightarrow\quad CF \:=\:\frac{4500\sin52.2^o}{\sin60.4^o} \:=\;4089.382738$
Therefore: .$\displaystyle \boxed{CF \:\approx\:4089}$