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

$\mathrm{\Delta }ABE$ is isosceles $\Rightarrow BD=DE=7$

$\mathrm{\Delta }ADC:\tan \left(90-2\theta \right)=\frac{AD}{21} ...\left(1\right)$

$\mathrm{\Delta }ADE:\tan \left(90-\theta \right)=\frac{AB}{7} ...\left(2\right)$

Divide $\frac{\tan \theta }{\tan 2\theta }=\frac{1}{3}\Rightarrow \frac{1-{\tan }^2\theta }{2}=\frac{1}{3}$

$1-\tan 2\theta =\frac{2}{3}\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}\Rightarrow \theta =30°$

$\mathrm{\Delta }ABD:\cos \left(90-\theta \right)=\frac{BD}{C}=\sin \theta$

$C=7cosec\theta =14$

$∆ADC:\cos \left(90-2\theta \right)=\frac{DC}{b}=\sin 2\theta$

$b=\frac{42}{\sqrt{3}}=14\sqrt{3}$

$[b+c]=38$