Question

$3tan{x}^{\circ }-2sin{y}^{\circ }+4cos{y}^{\circ }$ is $m\frac{2}{5}$, m is

Answer 1

Consider the bigger triangle as $△ABC$, such that, $\angle ABC={90}^{\circ }$
$AC=17$, $AB=8$ and a line from A meets BC at D, $BD=6$
Now, In $△ABC$,
Using Pythagoras Theorem,
$A{B}^2+B{C}^2=A{C}^2$
$8^2+B{C}^2={17}^2$
$B{C}^2=289-64$
$BC=15$
Now, In $△ABD$,
Using Pythagoras Theorem,
$A{B}^2+B{D}^2=A{D}^2$
$8^2+6^2=A{D}^2$
$A{D}^2=100$
$AD=10$
$3\tan x^{\circ }-2\sin y^{\circ }+4\cos y^{\circ }=3\left(\frac{P}{B}\right)-2\left(\frac{P}{H}\right)+4\left(\frac{B}{H}\right)$
= $3\left(\frac{AB}{BC}\right)-2\left(\frac{AB}{AD}\right)+4\left(\frac{BD}{AD}\right)$
= $3\left(\frac{8}{15}\right)-2\left(\frac{8}{10}\right)+4\left(\frac{6}{10}\right)$
= $\frac{8}{5}-\frac{8}{5}+\frac{12}{5}$
= $\frac{12}{5}=2\frac{2}{5}$