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}} $
$\Rightarrow \theta=30^{\circ}$
$\Delta ABD : \cos (90-\theta)=\frac{ BD }{ C }=\sin \theta$
$C =7 \operatorname{cosec} \theta=14$
$\Delta ADC : \cos (90-2 \theta)=\frac{ DC }{ b }=\sin 2 \theta$
$b =21 \operatorname{cosec} 2 \theta=21 \operatorname{cosec} \frac{\pi}{3}$
$b =\frac{42}{\sqrt{3}}=14 \sqrt{3}$
$b+c=14 \sqrt{3}+14 $
${[b+c]=38}$