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:3 - 2 cos x + 3 cot y is $5\frac{2}{m}$
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, $3-2\cos x+3\cot y$

= $3-2\left(\frac{AB}{AC}\right)+3\left(\frac{CD}{BD}\right)$
= $3-2\left(\frac{40}{50}\right)+3\left(\frac{24}{18}\right)$
= $3-2\left(\frac{4}{5}\right)+3\left(\frac{4}{3}\right)$

= $3-\frac{8}{5}+4$

= $\frac{27}{5}$

= $5\frac{2}{5}$

$\therefore m=5$