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:$2co{t}^{2}y-2cose{c}^{2}y+3$

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, $2{\cot }^{2}y-2cose{c}^{2}y+3$

= $2\left(\right(\frac{CD}{BD}{)}^2-\left(\frac{BC}{BD}{)}^2\right)+3$
= $2\left(\right(\frac{24}{18}{)}^2-\left(\frac{30}{18}{)}^2\right)+3$
= $2\left(\frac{-324}{324}\right)+3$

=$1$