Question

In the following figure : $AD\perp BC$, $AC=26$,
$CD=10$, $BC=42$,
$\angle DAC=x$ and $\angle B=y$, then

$\frac{6}{\cos x}-\frac{5}{\cos y}+8\tan y$ is $6\frac{1}{m}$, where the value of m is

Answer 1

Given: $△ABC$, $AD\perp BC$, $AC=26$, $CD=10$, $BC=42$
Now, In $△ACD$,
$A{C}^2=C{D}^2+A{D}^2$
${26}^2={10}^2+A{D}^2$
$A{D}^2=576$
$AD=24$
Now, In $△ABD$,
$BD=BC-CD$
$BD=42-10$
$BD=32$
$A{B}^2=B{D}^2+A{D}^2$
$A{B}^2={32}^2+{24}^2$
$A{B}^2=1600$
$AB=40$
Now, $\frac{6}{\cos x}-\frac{5}{\cos y}+8\tan y$
= $6\sec x-5secy+8\tan y$
= $6\left(\frac{H}{B}\right)-5\left(\frac{H}{B}\right)+8\left(\frac{P}{B}\right)$
= $6\left(\frac{AC}{AD}\right)-5\left(\frac{AB}{BD}\right)+8(\frac{AD}{BD}$
= $6\left(\frac{26}{24}\right)-5\left(\frac{40}{32}\right)+8\left(\frac{24}{32}\right)$
= $\frac{13}{2}-\frac{25}{4}+6$
= $\frac{26-25+24}{4}$
= $\frac{25}{4}$
= $6\frac{1}{4}$