Question

find x and y in each triangle. show your work.
1)
right triangle with 30° angle, one leg 3, other leg x, hypotenuse y
2)
right triangle with 60° angle, one leg (5sqrt{3}), other leg x, hypotenuse y
3)
right triangle with 30° angle, hypotenuse 22, one leg x, other leg y
4)
right triangle with 60° angle, one leg 18, other leg x, hypotenuse y

Explanation:

1.) Triangle 1 (30° right triangle)

Step1: Find x using tan(30°)

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

Step2: Find y using sin(30°)

$\sin(30^\circ) = \frac{3}{y} \implies y = \frac{3}{\sin(30^\circ)} = \frac{3}{\frac{1}{2}} = 6$

2.) Triangle 2 (60° right triangle)

Step1: Find x using tan(60°)

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

Step2: Find y using sin(60°)

$\sin(60^\circ) = \frac{5\sqrt{3}}{y} \implies y = \frac{5\sqrt{3}}{\sin(60^\circ)} = \frac{5\sqrt{3}}{\frac{\sqrt{3}}{2}} = 10$

3.) Triangle 3 (30° right triangle)

Step1: Find x using sin(30°)

$\sin(30^\circ) = \frac{x}{22} \implies x = 22 \times \sin(30^\circ) = 22 \times \frac{1}{2} = 11$

Step2: Find y using cos(30°)

$\cos(30^\circ) = \frac{y}{22} \implies y = 22 \times \cos(30^\circ) = 22 \times \frac{\sqrt{3}}{2} = 11\sqrt{3}$

4.) Triangle 4 (60° right triangle)

Step1: Find x using tan(60°)

$\tan(60^\circ) = \frac{18}{x} \implies x = \frac{18}{\tan(60^\circ)} = \frac{18}{\sqrt{3}} = 6\sqrt{3}$

Step2: Find y using sin(60°)

$\sin(60^\circ) = \frac{18}{y} \implies y = \frac{18}{\sin(60^\circ)} = \frac{18}{\frac{\sqrt{3}}{2}} = 12\sqrt{3}$

Answer:

1.) $x = 3\sqrt{3}$, $y = 6$
2.) $x = 5$, $y = 10$
3.) $x = 11$, $y = 11\sqrt{3}$
4.) $x = 6\sqrt{3}$, $y = 12\sqrt{3}$