Question

In the given figure, $\angle B={90}^{\circ }$ and $\angle ADB=x^{\circ }$. If $\tan x^{\circ }-\cos x^{\circ }+3\sin x^{\circ }$ is $3\frac{m}{15}$, then find $m$.

Answer 1

In $△ABD$,

$A{B}^2+B{D}^2=A{D}^2$ .....(Pythagoras Theorem)
${12}^2+B{D}^2={15}^2$
$B{D}^2=225-144$
$BD=9$
Now, $\tan x-\cos x+3\sin x$

= $\frac{P}{B}-\frac{B}{H}+3\frac{P}{H}$

= $\frac{AB}{BD}-\frac{BD}{AD}+3\frac{AB}{AD}$

= $\frac{12}{9}-\frac{9}{15}+3\frac{12}{15}$

= $\frac{4}{3}-\frac{3}{5}+\frac{12}{5}$

= $\frac{20-9+36}{15}$

= $\frac{47}{15}$

= $3\frac{2}{15}$

$\therefore m=2$