Question

In the following figure:
BD = 18, DC = 24, AC = 50, $\angle A=x$, $\angle BCD=y$ and $\angle ABC={90}^{\circ }=\angle BDC$.

Find:

2 tan x - sin y is $\frac{9}{m}$
value of m is

Answer 1

In the given figure, $\angle ABC=\angle BDC={90}^{\circ }$
In $△BDC$,
$B{C}^2=B{D}^2+C{D}^2$ (Pythagoras Theorem)
$B{C}^2={18}^2+{24}^2$
$BC=30$
In $△ABC$,
$A{C}^2=A{B}^2+B{C}^2$
${50}^2=A{B}^2+{30}^2$
$A{B}^2=1600$
$AB=40$
Now, $2\tan x-\sin y$

= $2\left(\frac{BC}{AB}\right)-\left(\frac{BD}{BC}\right)$
= $\frac{60}{40}-\frac{18}{30}$
= $\frac{3}{2}-\frac{3}{5}$

= $\frac{9}{10}$