Question

Use the given figure to find:

$\sin x^0$ is $\frac{8}{m}$, then '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,

$A{B}^2+B{D}^2=A{D}^2$

$8^2+6^2=A{D}^2$

$A{D}^2=100$

$AD=10$
$sin{x}^{\circ }=\frac{P}{H}=\frac{AB}{AC}=\frac{8}{17}$