Question

find the missing side lengths. leave your answers as radicals in simplest form.
13)
14)
15)
16)
17)
18)
19)

Explanation:

13)

Step1: Identify hypotenuse, solve for $x$

The side length 4 is the hypotenuse. Use cosine for $x$:
$\cos(30^\circ)=\frac{x}{4} \implies x=4\cos(30^\circ)=4\cdot\frac{\sqrt{3}}{2}=2\sqrt{3}$

Step2: Solve for $y$ using sine

$\sin(30^\circ)=\frac{y}{4} \implies y=4\sin(30^\circ)=4\cdot\frac{1}{2}=2$

14)

Step1: Solve for $y$ using tangent

$\tan(60^\circ)=\frac{y}{5} \implies y=5\tan(60^\circ)=5\sqrt{3}$

Step2: Solve for $x$ using cosine

$\cos(60^\circ)=\frac{5}{x} \implies x=\frac{5}{\cos(60^\circ)}=\frac{5}{\frac{1}{2}}=10$

15)

Step1: Solve for $x$ using cosine

$\cos(60^\circ)=\frac{8}{x} \implies x=\frac{8}{\cos(60^\circ)}=\frac{8}{\frac{1}{2}}=16$

Step2: Solve for $y$ using tangent

$\tan(60^\circ)=\frac{y}{8} \implies y=8\tan(60^\circ)=8\sqrt{3}$

16)

Step1: Identify 45-45-90 triangle, solve $y$

In a 45-45-90 triangle, legs are equal: $y=\frac{7\sqrt{2}}{2}$

Step2: Solve for hypotenuse $x$

$x=y\sqrt{2}=\frac{7\sqrt{2}}{2}\cdot\sqrt{2}=\frac{7\cdot2}{2}=7$

17)

Step1: Solve for $y$ using tangent

$\tan(30^\circ)=\frac{2}{y} \implies y=\frac{2}{\tan(30^\circ)}=\frac{2}{\frac{1}{\sqrt{3}}}=2\sqrt{3}$

Step2: Solve for $x$ using sine

$\sin(30^\circ)=\frac{2}{x} \implies x=\frac{2}{\sin(30^\circ)}=\frac{2}{\frac{1}{2}}=4$

18)

Step1: Identify 45-45-90 triangle, solve $y$

Legs are equal: $y=2\cdot\sin(45^\circ)=2\cdot\frac{\sqrt{2}}{2}=\sqrt{2}$

Step2: Solve for $x$

$x=y=\sqrt{2}$ (or $x=2\cdot\cos(45^\circ)=\sqrt{2}$)

19)

Step1: Identify 45-45-90 triangle, solve $y$

Legs are equal: $y=7$

Step2: Solve for hypotenuse $x$

$x=7\sqrt{2}$

Answer:

1. $x=2\sqrt{3}$, $y=2$
2. $x=10$, $y=5\sqrt{3}$
3. $x=16$, $y=8\sqrt{3}$
4. $x=7$, $y=\frac{7\sqrt{2}}{2}$
5. $x=4$, $y=2\sqrt{3}$
6. $x=\sqrt{2}$, $y=\sqrt{2}$
7. $x=7\sqrt{2}$, $y=7$