Solve this

Given: $\sin \theta=\frac{12}{13}$

Since, $\sin \theta=\frac{P}{H}$

$\Rightarrow P=12$ and $H=13$

Using Pythagoras theorem,

$P^{2}+B^{2}=H^{2}$

$\Rightarrow 12^{2}+B^{2}=13^{2}$

$\Rightarrow B^{2}=169-144$

$\Rightarrow B^{2}=25$

$\Rightarrow B=5$

Therefore,

$\cos \theta=\frac{B}{H}=\frac{5}{13}$

$\cos \theta=\frac{B}{H}=\frac{5}{13}$

Now,

$\left(\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}\right)=\left(\frac{2\left(\frac{12}{13}\right)-3\left(\frac{5}{13}\right)}{4\left(\frac{12}{13}\right)-9\left(\frac{5}{13}\right)}\right)$

$=\left(\frac{\frac{24}{13}-\frac{15}{13}}{\frac{48}{13}-\frac{45}{13}}\right)$

$=\left(\frac{24-15}{48-45}\right)$

$=\frac{9}{3}$

$=3$

Hence, $\left(\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}\right)=3$.