Question

Aplique a lei dos senos e determine as medidas desconhecidas no triângulo dado:

Answer 1

y/seno45=x/seno75°=20/seno60º

x/seno75°=20/seno60º

x/0,96593=20/08660

0,8660x=19,3186

x=19,3186/0,8660

x=22,30...

y/sen45=20/sen60

y/0,7071=20/0,8660

0,8660y=14,142

y=14,142/0,8660

y=16,33...

Answer 2

$\frac{y}{ \sin(45) } = \frac{20}{ \sin(60) } \\ \frac{y \sqrt{3} }{2} = 20 \times \frac{ \sqrt{2} }{2} \\ y = \frac{20 \sqrt{2} }{ \sqrt{3} } = \frac{20 \sqrt{6} }{3}$
$\frac{20}{ \sin(60) } = \frac{x}{ \sin(75) } \\ 20( \sin(45 + 30) ) = \frac{ x\sqrt{3} }{2} \\ 20( \sin(45) \times \cos(30) + \sin(30) \times \cos(45) ) = \frac{x \sqrt{3} }{2} \\ 20( \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} + \frac{1}{2} \times \frac{ \sqrt{2} }{2} ) = \frac{x \sqrt{3} }{2} \\ 20( \frac{ \sqrt{6} }{4} + \frac{ \sqrt{2} }{4} ) = \frac{x \sqrt{3} }{2} \\ 5( \sqrt{6} + \sqrt{2 } ) = \frac{x \sqrt{3} }{2} \\ 10( \sqrt{6} + \sqrt{2} ) = x \sqrt{3} \\ \frac{10( \sqrt{6} + \sqrt{2} )}{ \sqrt{3} } = x \\ \frac{10 \sqrt{3}( \sqrt{6} + \sqrt{2} ) }{3} = x \\ \frac{10 \sqrt{18} + 10 \sqrt{6} }{3} = x$