Question

Find the missing side lengths. Leave your answers as radicals in simplest form.

Answer 1

Answer:

A.

$x = \frac{7\sqrt{6}{3}$

$y = \frac{7\sqrt{6}}{3}$

Step-by-step explanation:

Reference angle = 60°

Opposite = $\frac{7\sqrt{2}}{2}$

Hypotenuse = x

Adjacent = y

✔️To find x, apply the trigonometric function SOH:

Sin 60° = Opp/Hyp

$sin 60° = \frac{\frac{7\sqrt{2}}{2}}{x}$

$\frac{\sqrt{3}}{2} = \frac{\frac{7\sqrt{2}}{2}}{x}$ (sin 60 = √3/2)

$\frac{\sqrt{3}}{2} = \frac{7\sqrt{2}}{2}*\frac{1}{x}$

$\frac{\sqrt{3}}{2} = \frac{7\sqrt{2}}{2x}$

Cross multiply

$\sqrt{3}*2x = 7\sqrt{2}*2$

$2\sqrt{3}*x = 14\sqrt{2}$

Divide both sides by 2

$\sqrt{3}*x = 7\sqrt{2}$

Divide both sides by √3

$x = \frac{7\sqrt{2}}{\sqrt{3}}$

Rationalize

$x = \frac{7\sqrt{2}*\sqrt{3}}{\sqrt{3}*\sqrt{3}}$

$x = \frac{7\sqrt{6}{3}$

✔️To find y, apply the trigonometric function TOA:

Tan 60° = Opp/Adjacent

$Tan 60° = \frac{\frac{7\sqrt{2}}{2}}{y}$

$\sqrt{3} = \frac{\frac{7\sqrt{2}}{2}}{y}$ (tan 60 = √3)

$\sqrt{3} = \frac{7\sqrt{2}}{2}*\frac{1}{y}$

$\sqrt{3} = \frac{7\sqrt{2}}{2y}$

Cross multiply

$\sqrt{3}*2y = 7\sqrt{2}$

$2\sqrt{3}*y = 7\sqrt{2}$

Divide both sides by √3

$y = \frac{7\sqrt{2}}{\sqrt{3}}$

Rationalize

$y = \frac{7\sqrt{2}*\sqrt{3}}{\sqrt{3}*\sqrt{3}}$

$y = \frac{7\sqrt{6}}{3}$