Question

Find y on this special right triangle

Answer 1

Answer:

• Cos(x) = adjacent side/hypotenuse

$cos(45)=\frac{\frac{7\sqrt{2} }{2} }{x} \\\\cos(45)x=\frac{7\sqrt{2} }{2}\\\\\frac{\sqrt{2} }{2}x=\frac{7\sqrt{2} }{2}\\\\x=\frac{\frac{7\sqrt{2} }{2}}{\frac{\sqrt{2} }{2}} =\frac{7\sqrt{2} }{2}*\frac{2}{\sqrt{2}} =7$

• Use the Pythagorean Theorem to find y

$y^{2} +(\frac{7\sqrt{2} }{2} )^{2} =x^{2} \\\\y^{2} =7^{2}-(\frac{7\sqrt{2} }{2} )^{2} \\\\y^{2}=49-\frac{49(2)}{4} =49-\frac{49}{2}=\frac{49(2)}{2}-\frac{49}{2}=\frac{98-49}{2}=\frac{49}{2} \\\\y=\sqrt{\frac{49}{2} } =\frac{7}{\sqrt{2} } =\frac{7\sqrt{2}}{\sqrt{2}(\sqrt{2})} =\frac{7\sqrt{2}}{2}$