Question

find the values of ( x ) and ( y ). (see example 2.)
4.
5.
6.

Explanation:

Problem 4

Step1: Identify trigonometric ratio for x

We use $\tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{3\sqrt{3}}{x}$
$\tan(60^\circ) = \sqrt{3}$, so $\sqrt{3} = \frac{3\sqrt{3}}{x}$

Step2: Solve for x

Rearrange to solve for x: $x = \frac{3\sqrt{3}}{\sqrt{3}} = 3$

Step3: Identify trigonometric ratio for y

Use $\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{y}$
$\cos(60^\circ) = \frac{1}{2}$, so $\frac{1}{2} = \frac{3}{y}$

Step4: Solve for y

Rearrange to solve for y: $y = 3 \times 2 = 6$

Problem 5

Step1: Identify trigonometric ratio for y

We use $\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{y}{24}$
$\cos(60^\circ) = \frac{1}{2}$, so $\frac{1}{2} = \frac{y}{24}$

Step2: Solve for y

Rearrange to solve for y: $y = 24 \times \frac{1}{2} = 12$

Step3: Identify trigonometric ratio for x

Use $\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{24}$
$\sin(60^\circ) = \frac{\sqrt{3}}{2}$, so $\frac{\sqrt{3}}{2} = \frac{x}{24}$

Step4: Solve for x

Rearrange to solve for x: $x = 24 \times \frac{\sqrt{3}}{2} = 12\sqrt{3}$

Problem 6

Step1: Identify trigonometric ratio for x

We use $\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{12\sqrt{3}}$
$\cos(30^\circ) = \frac{\sqrt{3}}{2}$, so $\frac{\sqrt{3}}{2} = \frac{x}{12\sqrt{3}}$

Step2: Solve for x

Rearrange to solve for x: $x = 12\sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{12 \times 3}{2} = 18$

Step3: Identify trigonometric ratio for y

Use $\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{12\sqrt{3}}$
$\sin(30^\circ) = \frac{1}{2}$, so $\frac{1}{2} = \frac{y}{12\sqrt{3}}$

Step4: Solve for y

Rearrange to solve for y: $y = 12\sqrt{3} \times \frac{1}{2} = 6\sqrt{3}$

Answer:

1. $x=3$, $y=6$
2. $x=12\sqrt{3}$, $y=12$
3. $x=18$, $y=6\sqrt{3}$