Question

If in a triangle ABC, $BC=5,CA=4,AB=3$ and D, E are points on BC such that $BD=DE=EC,\angle CAE=\theta$

• A. $A{E}^2=\frac{73}{3}$
• B. $A{E}^2=\frac{73}{9}$
• C. $\tan \theta =\frac{3}{8}$
• D. $\cos \theta =3\sqrt{73}$

Answer 1

Answer: (B), (C)

$A{E}^2=\frac{73}{9}$
$\tan \theta =\frac{3}{8}$

Hence
$BD=DE=EC=\frac{5}{3}$ and $△ABC$ is a right angled triangle.
Now
$cosC=\frac{A{C}^2+E{C}^2-A{E}^2}{2.AC.EC}$
Or
$A{E}^2=A{C}^2+E{C}^2-2.AC.EC.cosC$
Or
$=4^2+(\frac{5}{3}{)}^2-\mathrm{2.8.}\frac{5}{3}.cosC$
$=16+\frac{25}{9}-\frac{2\times 4\times 5}{3}.\frac{4}{5}$
Or
$=\frac{169}{9}-\frac{32}{3}$
$=\frac{169-96}{9}$
$=\frac{73}{9}$.
Hence
$A{E}^2=\frac{73}{9}$.
$cos\theta =\frac{A{E}^2+4^2-{\frac{5}{3}}^2}{2.AE.4}$
$=\frac{8}{\sqrt{73}}$.
Hence
$sec\theta =\frac{\sqrt{73}}{8}$
Now
$tan\theta =\sqrt{se{c}^2\theta -1}$
$=\sqrt{\frac{73}{64}-1}$
$=\frac{3}{8}$.