Question

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

Answer 1

A.

`y = (7√(6))/(3)`

Explanation:

Reference angle = 60°

Opposite =
`(7√(2))/(2)`

Hypotenuse = x

Adjacent = y

✔️To find x, apply the trigonometric function SOH:

Sin 60° = Opp/Hyp

`sin 60° = ((7√(2))/(2))/(x)`

`(√(3))/(2) = ((7√(2))/(2))/(x)` (sin 60 = √3/2)

`(√(3))/(2) = (7√(2))/(2)*(1)/(x)`

`(√(3))/(2) = (7√(2))/(2x)`

Cross multiply

`√(3)*2x = 7√(2)*2`

`2√(3)*x = 14√(2)`

Divide both sides by 2

`√(3)*x = 7√(2)`

Divide both sides by √3

`x = (7√(2))/(√(3))`

Rationalize

`x = (7√(2)*√(3))/(√(3)*√(3))`

✔️To find y, apply the trigonometric function TOA:

Tan 60° = Opp/Adjacent

`Tan 60° = ((7√(2))/(2))/(y)`

`√(3) = ((7√(2))/(2))/(y)` (tan 60 = √3)

`√(3) = (7√(2))/(2)*(1)/(y)`

`√(3) = (7√(2))/(2y)`

Cross multiply

`√(3)*2y = 7√(2)`

`2√(3)*y = 7√(2)`

Divide both sides by √3

`y = (7√(2))/(√(3))`

Rationalize

`y = (7√(2)*√(3))/(√(3)*√(3))`

`y = (7√(6))/(3)`