Question

fill in the blank 1 point
if c = 18√3, find b and c.
a= choose your answer... b= choose your answer...

Explanation:

Step1: Identify triangle type

This is 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 say length \(x\)), opposite \(60^\circ\) is \(x\sqrt{3}\), and hypotenuse is \(2x\). Here, hypotenuse \(c = 18\sqrt{3}\), side \(a\) is opposite \(30^\circ\)? Wait, no: angle \(30^\circ\) is at the base, so side \(a\) is opposite \(30^\circ\)? Wait, no: right angle is between \(a\) and \(b\), so angle \(30^\circ\) is opposite \(a\), angle \(60^\circ\) is opposite \(b\), hypotenuse \(c\). So in 30-60-90 triangle, side opposite \(30^\circ\) (a) is \( \frac{c}{2} \)? Wait no, hypotenuse is \(c\), so:

- Side opposite \(30^\circ\) (a) : \( \frac{c}{2} \)
- Side opposite \(60^\circ\) (b) : \( \frac{c\sqrt{3}}{2} \)? Wait no, wait the ratios: in 30-60-90, hypotenuse \(= 2 \times\) shorter leg (opposite 30°). Shorter leg (a, opposite 30°) : \( \frac{c}{2} \). Longer leg (b, opposite 60°) : \( a\sqrt{3} = \frac{c\sqrt{3}}{2} \)? Wait no, let's correct:

Wait, angle 30°: opposite side is \(a\), angle 60°: opposite side is \(b\), hypotenuse \(c\). So:

\(\sin(30^\circ) = \frac{a}{c}\) → \(a = c \sin(30^\circ) = c \times \frac{1}{2}\)

\(\sin(60^\circ) = \frac{b}{c}\) → \(b = c \sin(60^\circ) = c \times \frac{\sqrt{3}}{2}\)

Given \(c = 18\sqrt{3}\)

Step2: Calculate a

\(a = c \times \frac{1}{2} = 18\sqrt{3} \times \frac{1}{2} = \frac{18\sqrt{3}}{2} = 9\sqrt{3}\)? Wait no, wait that can't be. Wait, wait maybe I mixed up the angles. Wait the right angle is between \(a\) and \(b\), so angle at the bottom is 30°, so angle 30° is adjacent to \(b\) and opposite to \(a\). Wait, no: in the triangle, the right angle is at the bottom left, so:

- Vertical leg: \(a\) (adjacent to 30° angle, opposite 60°? Wait no, let's label the triangle:

- Right angle: vertex at bottom left, so:

- Top vertex: 60°, bottom right: 30°, bottom left: right angle.

- So sides:

- \(a\): vertical leg (left), adjacent to 30° angle, opposite 60° angle.

- \(b\): horizontal leg (bottom), adjacent to 60° angle, opposite 30° angle.

- \(c\): hypotenuse.

So angle 30°: opposite side is \(a\)? No, angle 30° is at bottom right, so the side opposite is \(a\) (vertical leg). Angle 60° is at top, opposite side is \(b\) (horizontal leg). Hypotenuse \(c\).

So in 30-60-90 triangle:

- Opposite 30° (a) : shorter leg, length \(x\)

- Opposite 60° (b) : longer leg, length \(x\sqrt{3}\)

- Hypotenuse (c) : \(2x\)

So hypotenuse \(c = 2x\), so \(x = \frac{c}{2}\). Then \(a = x = \frac{c}{2}\), \(b = x\sqrt{3} = \frac{c\sqrt{3}}{2}\)

Given \(c = 18\sqrt{3}\)

So \(a = \frac{18\sqrt{3}}{2} = 9\sqrt{3}\)? Wait no, wait that would make \(b = 9\sqrt{3} \times \sqrt{3} = 9 \times 3 = 27\). Wait, let's check with trigonometry:

\(\sin(30^\circ) = \frac{a}{c}\) → \(a = c \times \sin(30^\circ) = 18\sqrt{3} \times \frac{1}{2} = 9\sqrt{3}\)

\(\cos(30^\circ) = \frac{b}{c}\) → \(b = c \times \cos(30^\circ) = 18\sqrt{3} \times \frac{\sqrt{3}}{2} = 18 \times \frac{3}{2} = 27\)

Wait, that makes sense. Alternatively, using the ratio: hypotenuse \(c = 18\sqrt{3}\), so shorter leg (a, opposite 30°) is \( \frac{c}{2} = 9\sqrt{3} \), longer leg (b, opposite 60°) is \( a\sqrt{3} = 9\sqrt{3} \times \sqrt{3} = 27 \)

Step3: Wait, the problem says "find b and c" but c is given? Wait, maybe typo: "find a and b"? Because c is given as \(18\sqrt{3}\). Let's recheck the problem: "If c = 18√3, find b and c." Wait, that's a typo, probably "find a and b".

Assuming that, let's proceed:

So \(a = 9\sqrt{3}\), \(b = 27\)

Wait, let's confirm with Pythagoras: \(a^2 + b^2 = c^2\)

\( (9\sqrt{3})^2 + 27^2 = 81 \times 3 + 729 = 243 + 729 = 972 \)

\( c^2 = (18\sqrt{3})^2 = 324 \times 3 = 972 \). Correct.

So if the problem is to find a and b (since c is given), then:

a = \(9\sqrt{3}\), b = 27

Answer:

a = \(9\sqrt{3}\), b = 27