Question

If a, b, c the sides of a triangle ABC be $5,4,3$ respectively and D, E are the points of trisection of side BC, then prove that $tan\angle CAE=3/8$.

Answer 1

$5^2=4^2+3^2\therefore \angle A={90}^0$,
$\therefore cosC=4/5$
Also $CE=\frac{1}{3}CB=\frac{5}{3}$
Hence by cosine formula on $△AEC$
$A{E}^2=4^2+(\frac{5}{3}{)}^2-\mathrm{2.4.}\frac{5}{3}cosC$
$=\frac{169}{9}-\frac{40}{3}.\frac{4}{5}=\frac{73}{9}$
Now we know all the sides of $△ACE$
$\therefore$ By cosine rule, $cos\theta =\frac{8}{\sqrt{73}}\therefore tan\theta =\frac{3}{8}$.