Question

find the missing side lengths. leave your answers as radicals in simplest form.
1)
2)
3)
4)
5)
6)
7)
8)

Explanation:

1)

Step1: Identify 45-45-90 triangle properties

In a 45-45-90 triangle, legs are equal, hypotenuse = leg $\times\sqrt{2}$. Given leg $=5\sqrt{3}$, so $y=5\sqrt{3}$.

Step2: Calculate hypotenuse $x$

$x = 5\sqrt{3} \times \sqrt{2} = 5\sqrt{6}$

2)

Step1: Find side $m$ (opposite 60°)

$\sin60^\circ=\frac{m}{4}$, so $m=4\times\frac{\sqrt{3}}{2}=2\sqrt{3}$

Step2: Find side $n$ (adjacent 60°)

$\cos60^\circ=\frac{n}{4}$, so $n=4\times\frac{1}{2}=2$

3)

Step1: Identify 45-45-90 triangle

Legs are equal, so $y=\sqrt{2}$

Step2: Calculate hypotenuse $x$

$x = \sqrt{2} \times \sqrt{2} = 2$

4)

Step1: Find side $b$ (opposite 30°)

$\tan30^\circ=\frac{b}{7\sqrt{3}}$, so $b=7\sqrt{3}\times\frac{1}{\sqrt{3}}=7$

Step2: Find hypotenuse $a$

$\cos30^\circ=\frac{7\sqrt{3}}{a}$, so $a=\frac{7\sqrt{3}}{\frac{\sqrt{3}}{2}}=14$

5)

Step1: Identify 45-45-90 triangle

Legs are equal, so $n=5$

Step2: Calculate hypotenuse $m$

$m = 5 \times \sqrt{2} = 5\sqrt{2}$

6)

Step1: Find side $y$ (adjacent 60°)

$\cos60^\circ=\frac{y}{6}$, so $y=6\times\frac{1}{2}=3$

Step2: Find side $x$ (opposite 60°)

$\sin60^\circ=\frac{x}{6}$, so $x=6\times\frac{\sqrt{3}}{2}=3\sqrt{3}$

7)

Step1: Identify 45-45-90 triangle

Legs are equal, so $x=5\sqrt{2}$

Step2: Calculate hypotenuse $w$

$w = 5\sqrt{2} \times \sqrt{2} = 10$

8)

Step1: Find side $y$ (opposite 60°)

$\tan60^\circ=\frac{y}{8}$, so $y=8\times\sqrt{3}=8\sqrt{3}$

Step2: Find hypotenuse $x$

$\cos60^\circ=\frac{8}{x}$, so $x=\frac{8}{\frac{1}{2}}=16$

Answer:

1. $x=5\sqrt{6}$, $y=5\sqrt{3}$
2. $m=2\sqrt{3}$, $n=2$
3. $x=2$, $y=\sqrt{2}$
4. $a=14$, $b=7$
5. $m=5\sqrt{2}$, $n=5$
6. $x=3\sqrt{3}$, $y=3$
7. $x=5\sqrt{2}$, $w=10$
8. $x=16$, $y=8\sqrt{3}$