Question

$\tan \angle DBC$ is $\frac{3}{m}$
value of m is

Answer 1

In $△ABC$
$A{B}^2+B{C}^2=A{C}^2$ (By Pythagoras theorem)
$4^2+3^2=A{C}^2$
$A{C}^2=25$
$AC=5$
In $△ABC$
$\angle A+\angle B+\angle C=180$
$\angle A+\angle C=90$ (As $\angle B=90$).....(i)
In $△ABD$,
$\angle A+\angle BDA+\angle DBA=180$
$\angle A+\angle DBA=90$ (As $\angle BDA=90$).....(i)
From (i) and (ii)
$\angle C=\angle DBA$
Similarly, $\angle DBC=\angle A$
Now, $\tan \angle DBC=\tan \angle A=\frac{P}{B}=\frac{BC}{AB}=\frac{3}{4}$