Question

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

$\cot x$ is $2.4$,

If true then write 1 and if false then write 0

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 $△ABD$,
$BD=AC-CD$
$BD=42-10$
$BD=32$
$A{B}^2=B{D}^2+A{D}^2$
$A{B}^2={32}^2+{24}^2$
$A{B}^2=1600$
$AB=40$
Now, $\cot x=\frac{B}{P}=\frac{AD}{CD}=\frac{24}{10}=2.4$