Find the value of the following from the figure given:
$\tan \theta$

Firstly we give the name to the midpoint of $BC$ ie., $M$
$BC=BM+MC=2BM$ or $2MC$
$\Rightarrow$ $BM=5,MC=5$
Now we have to find the value of $AM$ with the help of Pythagoras theorem
So, In $△AMB$
$\Rightarrow {\left(AM\right)}^2+{\left(BM\right)}^2={\left(AB\right)}^2\Rightarrow {\left(AM\right)}^2+{\left(5\right)}^2={\left(13\right)}^2\Rightarrow {\left(AM\right)}^2={\left(13\right)}^2-{\left(5\right)}^2$
using the identity $a^2-b^2=(a+b)(a-b)$
$\Rightarrow {\left(AM\right)}^2=(13-5)(13+5)\Rightarrow {\left(AM\right)}^2=(8)(18)=144$
$\Rightarrow AM=\sqrt{144}=\pm 12$
But side $AM$ can't be negative. So, $AM=12$
$\tan \theta$
we know that
$\tan \theta =\frac{sideoppositetoangle\theta }{sideadjacenttoangle\theta }$
In $△AMB$
side opposite to $\theta =AM=12$
side adjacent to $\theta =BM=5$
so,
$\tan \theta =\frac{AM}{BM}=\frac{12}{5}$