y = 8√3 simplify.
find the values of x and y. give your answers in simplest radical form.
5.
triangle with right angle, 30° angle, 60° angle, hypotenuse? labeled y, x, and base 18
6.
triangle with right angle, 30° angle, 60° angle, labeled 2, x, y
7.
triangle with right angle, 60° angle, 30° angle, labeled x, y, and base 24√3
8.
triangle with right angle, 30° angle, 60° angle, labeled y, x, and base 33

Question

Accepted Answer

### Problem 5:
# Explanation:
## Step1: Identify triangle type
It's a 30-60-90 right triangle. In such a triangle, the sides are in the ratio $1 : \sqrt{3} : 2$, where the side opposite $30^\circ$ is the shortest (let's call it $a$), opposite $60^\circ$ is $a\sqrt{3}$, and hypotenuse is $2a$. Here, the hypotenuse is $18$ (opposite the right angle? Wait, no: the hypotenuse is the side opposite the right angle. Wait, the angles are $30^\circ$, $60^\circ$, and $90^\circ$. The side opposite $30^\circ$ is $x$ (wait, no: looking at the triangle, the right angle is between $x$ and $y$, the base is $18$ (hypotenuse? Wait, no, the hypotenuse is the longest side, opposite the right angle. Wait, the triangle has angles $30^\circ$, $60^\circ$, $90^\circ$, so the hypotenuse is the side opposite $90^\circ$, which is the base with length $18$. So hypotenuse $c = 18$. Then:

- Side opposite $30^\circ$ (let's say $x$ or $y$): Wait, the angle $30^\circ$ is at the left, $60^\circ$ at the right, right angle at the top. So the side opposite $30^\circ$ is $x$ (vertical side), opposite $60^\circ$ is $y$ (horizontal side), and hypotenuse is $18$ (base).

In 30-60-90 triangle:

- Hypotenuse $c = 2a$, where $a$ is the side opposite $30^\circ$ (shortest side).

- Side opposite $60^\circ$ is $a\sqrt{3}$.

So hypotenuse $c = 18 = 2a \implies a = 9$. Wait, no: wait, the side opposite $30^\circ$ is the shortest, so if hypotenuse is $18$, then $a = \frac{18}{2} = 9$ (side opposite $30^\circ$, which is $x$? Wait, no, the angle at the right is $60^\circ$, so the side opposite $60^\circ$ is $y$? Wait, maybe I mixed up. Let's reorient:

Right angle at top, $30^\circ$ at left, $60^\circ$ at right. So:

- Angle at left: $30^\circ$, so the side opposite is $x$ (vertical side, adjacent to $60^\circ$ angle).

- Angle at right: $60^\circ$, side opposite is $y$ (horizontal side, adjacent to $30^\circ$ angle).

- Hypotenuse: base, length $18$, opposite right angle.

So in 30-60-90:

- Side opposite $30^\circ$ (x) = $\frac{hypotenuse}{2} = \frac{18}{2} = 9$? Wait, no, that can't be. Wait, no: the side opposite $30^\circ$ is the shortest, so if hypotenuse is $18$, then side opposite $30^\circ$ (x) is $9$, side opposite $60^\circ$ (y) is $9\sqrt{3}$? Wait, no, that seems reversed. Wait, no: let's recall the ratios correctly. In a 30-60-90 triangle, the sides are in the ratio $1 : \sqrt{3} : 2$, where:

- Shortest side (opposite $30^\circ$): $a$

- Side opposite $60^\circ$: $a\sqrt{3}$

- Hypotenuse: $2a$

So hypotenuse is $2a$, so if hypotenuse is $18$, then $2a = 18 \implies a = 9$. Then:

- Side opposite $30^\circ$ (x) = $a = 9$

- Side opposite $60^\circ$ (y) = $a\sqrt{3} = 9\sqrt{3}$

Wait, but let's check: if $x = 9$, $y = 9\sqrt{3}$, hypotenuse $18$, then $9^2 + (9\sqrt{3})^2 = 81 + 243 = 324 = 18^2$, which works. So that's correct.

## Step2: Solve for x and y
- Hypotenuse $c = 18 = 2a \implies a = 9$ (side opposite $30^\circ$, which is $x$)
- Side opposite $60^\circ$ (y) = $a\sqrt{3} = 9\sqrt{3}$

# Answer (Problem 5):
$x = 9$, $y = 9\sqrt{3}$

### Problem 6:
# Explanation:
## Step1: Identify triangle type
30-60-90 right triangle. Angles: $30^\circ$ (top right), $60^\circ$ (bottom left), $90^\circ$ (top left). The side adjacent to $60^\circ$ and opposite $30^\circ$ is $2$ (vertical side). Let's denote:

- Side opposite $30^\circ$ (angle at top right) is $2$ (vertical side, length $2$). Wait, angle at top right is $30^\circ$, so side opposite $30^\circ$ is the vertical side (length $2$)? Wait, no: angle at top right is $30^\circ$, so the side opposite $30^\circ$ is the side opposite to it, which is the vertical side (length $2$)? Wait, the triangle has right angle at top left, so the sides:

- Vertical side: length $2$ (adjacent to $60^\circ$, opposite $30^\circ$)
- Horizontal side: $x$ (adjacent to $30^\circ$, opposite $60^\circ$)
- Hypotenuse: $y$ (opposite right angle)

In 30-60-90 triangle:

- Side opposite $30^\circ$ (let's call it $a$) = $2$ (vertical side)
- Then, side opposite $60^\circ$ (x) = $a\sqrt{3} = 2\sqrt{3}$
- Hypotenuse (y) = $2a = 4$

Wait, let's verify:

- $a = 2$ (side opposite $30^\circ$)
- Side opposite $60^\circ$ (x) = $a\sqrt{3} = 2\sqrt{3}$
- Hypotenuse (y) = $2a = 4$

Check with Pythagoras: $2^2 + (2\sqrt{3})^2 = 4 + 12 = 16 = 4^2$. Correct.

## Step2: Calculate x and y
- $a = 2$ (side opposite $30^\circ$)
- $x = a\sqrt{3} = 2\sqrt{3}$
- $y = 2a = 4$

# Answer (Problem 6):
$x = 2\sqrt{3}$, $y = 4$

### Problem 7:
# Explanation:
## Step1: Identify triangle type
30-60-90 right triangle. Angles: $60^\circ$ (bottom left), $30^\circ$ (bottom right), $90^\circ$ (top left). The base (opposite $90^\circ$) is $24\sqrt{3}$ (hypotenuse? Wait, no: hypotenuse is opposite $90^\circ$, so the base is hypotenuse? Wait, the triangle has right angle at top left, so the base (bottom side) is hypotenuse, length $24\sqrt{3}$. Angles: $60^\circ$ at bottom left, $30^\circ$ at bottom right. So:

- Side opposite $30^\circ$ (y, vertical side) = $a$
- Side opposite $60^\circ$ (x, horizontal side) = $a\sqrt{3}$
- Hypotenuse (base) = $2a = 24\sqrt{3}$

## Step2: Solve for a
$2a = 24\sqrt{3} \implies a = 12\sqrt{3}$ (side opposite $30^\circ$, which is $y$)

- Side opposite $60^\circ$ (x) = $a\sqrt{3} = 12\sqrt{3} \times \sqrt{3} = 12 \times 3 = 36$

Wait, let's check:

- $a = 12\sqrt{3}$ (y)
- $x = 36$
- Hypotenuse: $2a = 24\sqrt{3}$, which matches.

Check Pythagoras: $x^2 + y^2 = 36^2 + (12\sqrt{3})^2 = 1296 + 432 = 1728$. Hypotenuse squared: $(24\sqrt{3})^2 = 576 \times 3 = 1728$. Correct.

# Answer (Problem 7):
$x = 36$, $y = 12\sqrt{3}$

### Problem 8:
# Explanation:
## Step1: Identify triangle type
30-60-90 right triangle. Angles: $30^\circ$ (bottom left), $60^\circ$ (top right), $90^\circ$ (bottom right). The base (adjacent to $30^\circ$) is $33$ (horizontal side), vertical side is $x$ (opposite $30^\circ$), hypotenuse is $y$ (opposite $90^\circ$).

In 30-60-90 triangle:

- Side adjacent to $30^\circ$ (let's call it $b$) = $33$ (this is the side opposite $60^\circ$, which is $b = a\sqrt{3}$, where $a$ is side opposite $30^\circ$ (x))
- Side opposite $30^\circ$ (x) = $a$
- Hypotenuse (y) = $2a$

So, $b = a\sqrt{3} = 33 \implies a = \frac{33}{\sqrt{3}} = 11\sqrt{3}$ (rationalizing the denominator: $\frac{33}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{33\sqrt{3}}{3} = 11\sqrt{3}$)

- Hypotenuse $y = 2a = 2 \times 11\sqrt{3} = 22\sqrt{3}$? Wait, no: wait, the side adjacent to $30^\circ$ is $33$, which is the side opposite $60^\circ$, so $b = a\sqrt{3} = 33 \implies a = \frac{33}{\sqrt{3}} = 11\sqrt{3}$ (x). Then hypotenuse $y = 2a = 22\sqrt{3}$? Wait, no, wait: let's re-express.

Wait, the angle at bottom left is $30^\circ$, right angle at bottom right, so:

- Adjacent to $30^\circ$: $33$ (horizontal side)
- Opposite to $30^\circ$: $x$ (vertical side)
- Hypotenuse: $y$ (slanted side)

In 30-60-90 triangle:

- $\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{33} \implies x = 33 \tan(30^\circ) = 33 \times \frac{1}{\sqrt{3}} = 11\sqrt{3}$
- $\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{33}{y} \implies y = \frac{33}{\cos(30^\circ)} = \frac{33}{\frac{\sqrt{3}}{2}} = \frac{66}{\sqrt{3}} = 22\sqrt{3}$

Wait, but let's check with the ratio. Alternatively, in 30-60-90, the sides are $a$ (opposite $30^\circ$), $a\sqrt{3}$ (opposite $60^\circ$), $2a$ (hypotenuse). Here, the side opposite $60^\circ$ is $33$ (since angle at top right is $60^\circ$, so side opposite $60^\circ$ is the adjacent side to $30^\circ$, which is $33$). So $a\sqrt{3} = 33 \implies a = \frac{33}{\sqrt{3}} = 11\sqrt{3}$ (x, opposite $30^\circ$). Then hypotenuse $y = 2a = 22\sqrt{3}$? Wait, no, that can't be, because $22\sqrt{3} \approx 38.1$, but $33$ is adjacent, and $x \approx 19.05$. Wait, maybe I mixed up the angles. Wait, the angle at top right is $60^\circ$, so the side opposite $60^\circ$ is the adjacent side to $30^\circ$ (length $33$), so $a\sqrt{3} = 33 \implies a = 11\sqrt{3}$ (x, opposite $30^\circ$), and hypotenuse $y = 2a = 22\sqrt{3}$. Wait, but let's check with Pythagoras:

$x^2 + 33^2 = y^2$

$(11\sqrt{3})^2 + 33^2 = (363) + (1089) = 1452$

$(22\sqrt{3})^2 = 484 \times 3 = 1452$. Correct.

# Answer (Problem 8):
$x = 11\sqrt{3}$, $y = 22\sqrt{3}$

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QUESTION IMAGE

Question

Explanation:

Problem 5:

Step1: Identify triangle type

Step2: Solve for x and y

Step1: Identify triangle type

Step2: Calculate x and y

Step1: Identify triangle type

Step2: Solve for a

Answer:

Problem 6: