Question

In a triangle $ABC$, the altitude $AD$ and the median $AE$ divide $\angle A$ into three equal parts. If $BC=28$, then the nearest integer to $AB+AC$ is

• A. 38
• B. 37
• C. 36
• D. 33

Answer 1

Answer: (A)

$\Delta ABE$ is isosceles $\Rightarrow BD=DE=7$
$\Delta ADC:\tan (90-2\theta )=\frac{AD}{21}$ ..... (1)
$\Delta ADE:\tan (90-\theta )=\frac{AB}{7}$ .....(2)
Divide $\frac{\tan \theta }{\tan 2\theta }=\frac{1}{3}$
$\Rightarrow \frac{1-{\tan }^2\theta }{2}=\frac{1}{3}$
$1-\tan 2\theta =\frac{2}{3}$
$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}$
$\Rightarrow \theta =30^{\circ }$
$\Delta ABD:\cos (90-\theta )=\frac{BD}{C}=\sin \theta$
$C=7\mathrm{cosec}\theta =14$
$\Delta ADC:\cos (90-2\theta )=\frac{DC}{b}=\sin 2\theta$
$b=21\mathrm{cosec}2\theta =21\mathrm{cosec}\frac{\pi }{3}$
$b=\frac{42}{\sqrt{3}}=14\sqrt{3}$
$b+c=14\sqrt{3}+14$
$[b+c]=38$