Question

find x on this special right triangle, 45 is not an option!!!!

Answer 1

let the line between 2 tria be y

sin 60/8√2 = sin 90/y

y=13.06

sin 45/13.06 = sin 90/x

x=18.46

Answer 2

Answer:

First, find the hypotenuse of the right triangle with the 60° & 30°.

• Hypotenuse = h
• sin(x) = opposite side/hypotenuse

$sin(60) = \frac{8\sqrt{2}}{h} \\\\sin(60)h=8\sqrt{2}\\\\\frac{\sqrt{3}}{2} h=8\sqrt{2}\\\\h=\frac{8\sqrt{2}}{\frac{\sqrt{3}}{2}}=8\sqrt{2}*\frac{2}{\sqrt{3}} =\frac{16\sqrt{2} }{\sqrt{3}} =\frac{16\sqrt{2}(\sqrt{3}) }{\sqrt{3}(\sqrt{3})} =\frac{16\sqrt{6} }{3}$

Use that side length to find x.

• sin(x) = opposite side/hypotenuse

$sin(45)=\frac{\frac{16\sqrt{6}}{3}}{x}\\\\sin(45)x=\frac{16\sqrt{6}}{3} \\\\\frac{\sqrt{2}}{2}x=\frac{16\sqrt{6}}{3} \\\\x=\frac{\frac{16\sqrt{6}}{3}}{\frac{\sqrt{2}}{2}}=\frac{16\sqrt{6}}{3}*\frac{2}{\sqrt{2}}=\frac{16\sqrt{2}\sqrt{3}(2)}{3\sqrt{2} }=\frac{32\sqrt{3} }{3}$