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:$5-3ta{n}^{2}x+3se{c}^{2}x$

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, $5-3{\tan }^{2}x+3{\sec }^{2}y$
= $5-3({\tan }^{2}x-{\sec }^{2}x)$
= $5-3\left(\right(\frac{BC}{AB}{)}^2-\left(\frac{AC}{AB}{)}^2\right)$
= $5-3\left(\frac{B{C}^2-A{C}^2}{A{B}^2}\right)$
= $5-3\left(\frac{{30}^2-{50}^2}{{40}^2}\right)$
= $5-3\left(\frac{-1600}{1600}\right)$
= $5+3$
= $8$