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

Given:

A figure.

To find:

The values of x and y.

Solution:

We know that, in a right angled triangle,

[tex]\sin \theta =\dfrac{Perpendicular}{Hypotenuse}[/tex]

[tex]\sin 60^\circ =\dfrac{x}{16}[/tex]

[tex]\dfrac{\sqrt{3}}{2} =\dfrac{x}{16}[/tex]

Multiply both sides by 16.

[tex]\dfrac{\sqrt{3}}{2}\times 16=\dfrac{x}{16}\times 16[/tex]

[tex]8\sqrt{3}=x[/tex]

And,

[tex]\cos \theta =\dfrac{Base}{Hypotenuse}[/tex]

[tex]\cos 60^\circ =\dfrac{y}{16}[/tex]

[tex]\dfrac{1}{2}=\dfrac{y}{16}[/tex]

Multiply both sides by 16.

[tex]\dfrac{16}{2}=y[/tex]

[tex]8=y[/tex]

Therefore, the missing side lengths are [tex]x=8\sqrt{3}[/tex] and [tex]y=8[/tex].

What's the answer to this?