Question

1. diagram: right triangle with right angle at top - left, vertical leg (left) labeled 12, 30° angle at bottom vertex, horizontal leg (top) labeled y, hypotenuse (right) labeled x

Explanation:

Step1: Find \( x \) using cosine

In a right triangle, \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \). Here, \( \theta = 30^\circ \), adjacent side is 12, hypotenuse is \( x \). So \( \cos30^\circ=\frac{12}{x} \). Since \( \cos30^\circ=\frac{\sqrt{3}}{2} \), we have \( \frac{\sqrt{3}}{2}=\frac{12}{x} \). Solving for \( x \), \( x=\frac{12\times2}{\sqrt{3}}=\frac{24}{\sqrt{3}} = 8\sqrt{3}\times\sqrt{3}/\sqrt{3}=8\sqrt{3}? \) Wait, no, rationalize: \( \frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3} \)? Wait, no, \( \cos30^\circ=\frac{\sqrt{3}}{2} \), so \( x=\frac{12}{\cos30^\circ}=\frac{12}{\frac{\sqrt{3}}{2}} = 12\times\frac{2}{\sqrt{3}}=\frac{24}{\sqrt{3}} = 8\sqrt{3}\times\sqrt{3}/\sqrt{3} \)? No, correct rationalization: multiply numerator and denominator by \( \sqrt{3} \), so \( \frac{24\sqrt{3}}{3}=8\sqrt{3} \)? Wait, no, 24 divided by 3 is 8, so \( 8\sqrt{3} \)? Wait, no, maybe I mixed up. Wait, the angle is 30 degrees, the adjacent side to 30 degrees is 12, hypotenuse is \( x \). Alternatively, in a 30-60-90 triangle, the sides are in ratio \( 1:\sqrt{3}:2 \), where the side opposite 30 is shortest, adjacent to 30 (longer leg) is \( \sqrt{3} \) times the shortest leg, hypotenuse is twice the shortest leg. Wait, here the side of length 12 is adjacent to 30 degrees, so it's the longer leg (opposite 60 degrees). So longer leg = \( \sqrt{3} \times \) shorter leg, hypotenuse = \( 2 \times \) shorter leg. Let shorter leg be \( y \) (wait, no, \( y \) is the other leg, opposite 30 degrees). Wait, the right angle is at the top, so the legs are 12 (vertical) and \( y \) (horizontal), hypotenuse \( x \). The angle at the bottom is 30 degrees, so the angle between the vertical leg (12) and hypotenuse \( x \) is 30 degrees. So:

- \( \cos(30^\circ)=\frac{\text{adjacent (vertical leg)}}{\text{hypotenuse}}=\frac{12}{x} \)
- \( \sin(30^\circ)=\frac{\text{opposite (horizontal leg } y\text{)}}{\text{hypotenuse}}=\frac{y}{x} \)

We know \( \cos30^\circ=\frac{\sqrt{3}}{2} \), so \( x=\frac{12}{\cos30^\circ}=\frac{12}{\frac{\sqrt{3}}{2}} = 12\times\frac{2}{\sqrt{3}}=\frac{24}{\sqrt{3}} = 8\sqrt{3}\times\sqrt{3}/\sqrt{3} \)? No, correct: \( \frac{24}{\sqrt{3}} = 8\sqrt{3} \) (since \( 24\div3=8 \), \( \sqrt{3}\times\sqrt{3}=3 \)). Wait, no, \( 24/\sqrt{3} = 8\sqrt{3} \) because \( \sqrt{3}\times8\sqrt{3}=24 \). Yes.

For \( y \), \( \sin30^\circ=\frac{y}{x} \), and \( \sin30^\circ=\frac{1}{2} \), so \( y = x\times\frac{1}{2} = 8\sqrt{3}\times\frac{1}{2}=4\sqrt{3} \)? Wait, no, that can't be. Wait, maybe I mixed up the angle. Wait, the right angle is at the top, so the angle at the bottom is 30 degrees, so the vertical leg is adjacent to 30 degrees, horizontal leg is opposite to 30 degrees. So:

\( \cos(30^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{12}{x} \implies x=\frac{12}{\cos30^\circ}=\frac{12}{\frac{\sqrt{3}}{2}} = 8\sqrt{3} \) (wait, 12*2=24, 24/sqrt(3)=8*3/sqrt(3)=8*sqrt(3), yes).

\( \tan(30^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{y}{12} \implies y = 12\times\tan(30^\circ)=12\times\frac{1}{\sqrt{3}} = 4\sqrt{3} \).

Alternatively, using 30-60-90 triangle ratios: in a 30-60-90 triangle, the sides are in the ratio \( 1 : \sqrt{3} : 2 \), where the side opposite 30° is the shortest (let's call it \( a \)), the side opposite 60° is \( a\sqrt{3} \), and the hypotenuse is \( 2a \). Here, the side opposite 60° is 12 (since the angle at the bottom is 30°, so the other acute angle is 60°), so \( a\sqrt{3}=12 \implies a=\frac{12}{\sqrt{3}} = 4\sqrt{3} \). Then the hypotenuse \( x = 2a = 8\sqrt{3} \), and the side opposite 30° (which is \( y \)) is \( a = 4\sqrt{3} \).

So:

- \( x = 8\sqrt{3} \) (or approximately 13.856)
- \( y = 4\sqrt{3} \) (or approximately 6.928)

Assuming we need to find \( x \) and \( y \), let's confirm:

For \( x \) (hypotenuse):

\( \cos(30^\circ) = \frac{12}{x} \implies x = \frac{12}{\cos(30^\circ)} = \frac{12}{\frac{\sqrt{3}}{2}} = \frac{24}{\sqrt{3}} = 8\sqrt{3} \approx 13.86 \)

For \( y \) (opposite 30°):

\( \sin(30^\circ) = \frac{y}{x} \implies y = x \times \sin(30^\circ) = 8\sqrt{3} \times \frac{1}{2} = 4\sqrt{3} \approx 6.93 \)

Or using tangent:

\( \tan(30^\circ) = \frac{y}{12} \implies y = 12 \times \tan(30^\circ) = 12 \times \frac{1}{\sqrt{3}} = 4\sqrt{3} \)

Yes, that's correct.

Answer:

If finding \( x \): \( \boldsymbol{8\sqrt{3}} \) (or approximately 13.86)
If finding \( y \): \( \boldsymbol{4\sqrt{3}} \) (or approximately 6.93)

(Assuming the problem is to find \( x \) and \( y \) in the right triangle with a 30° angle and one leg 12. The above steps show the calculations for both sides.)