Question

solve for x.
14
x°
13
83°
x = ?°
law of sines: \\(\frac{\sin a}{a} = \frac{\sin b}{b} = \frac{\sin c}{c}\\)
round your answer to the nearest hundredth.

Explanation:

Step1: Identify sides and angles

Let’s denote the triangle with sides: \( a = 13 \), \( b = 14 \), angle opposite \( a \) is \( 83^\circ \), angle opposite \( b \) is \( x^\circ \).

Step2: Apply Law of Sines

Using \(\frac{\sin A}{a}=\frac{\sin B}{b}\), substitute \( A = 83^\circ \), \( a = 13 \), \( b = 14 \):
\(\frac{\sin 83^\circ}{13}=\frac{\sin x}{14}\)

Step3: Solve for \(\sin x\)

\(\sin x=\frac{14\times\sin 83^\circ}{13}\)
Calculate \(\sin 83^\circ\approx0.9925\), so \(\sin x=\frac{14\times0.9925}{13}\approx\frac{13.895}{13}\approx1.0688\)
Wait, this can't be right. Wait, maybe I mixed up the sides. Wait, the side opposite \( 83^\circ \) is 13? Wait no, the side adjacent to \( 83^\circ \) is 13? Wait, no, the triangle: the side of length 13 is opposite the angle \( x \)? Wait, no, let's re - examine the triangle. The side with length 14 is opposite the \( 83^\circ \) angle? Wait, no, the angle of \( 83^\circ \) has a side of length 13 adjacent? Wait, no, let's label correctly. Let's say: angle \( C = 83^\circ \), side \( c = 13 \), side \( b = 14 \), angle \( B=x \), side \( a \) is the other side. Wait, maybe I had the sides reversed. Let's correct: Let the angle \( 83^\circ \) be opposite side 14? No, the diagram: the side of length 13 is the base, the side of length 14 is one of the other sides, and the angle of \( 83^\circ \) is at the base - side and the side of length 14? Wait, no, let's use the Law of Sines correctly. Let's assume: angle \( A = x \), side \( a = 14 \); angle \( B = 83^\circ \), side \( b = 13 \). Then \(\frac{\sin A}{a}=\frac{\sin B}{b}\)
So \(\frac{\sin x}{14}=\frac{\sin 83^\circ}{13}\)
Then \(\sin x=\frac{14\times\sin 83^\circ}{13}\)
\(\sin 83^\circ\approx0.992546\)
\(14\times0.992546 = 13.895644\)
\(\sin x=\frac{13.895644}{13}\approx1.0689\)
Wait, this is greater than 1, which is impossible. So I must have mixed up the sides. Ah! Wait, the side of length 13 is opposite the angle \( x \), and the side of length 14 is opposite the \( 83^\circ \) angle. So let's re - define: angle \( A=x \), side \( a = 13 \); angle \( B = 83^\circ \), side \( b = 14 \)
Then \(\frac{\sin x}{13}=\frac{\sin 83^\circ}{14}\)
\(\sin x=\frac{13\times\sin 83^\circ}{14}\)
\(\sin 83^\circ\approx0.992546\)
\(13\times0.992546 = 12.903098\)
\(\sin x=\frac{12.903098}{14}\approx0.9216\)

Step4: Find \( x \)

\(x=\arcsin(0.9216)\approx67.2^\circ\) (Wait, but let's check the triangle sum. Wait, maybe the first mistake was in side - angle correspondence. Let's do it again. Let's label the triangle: let the angle at the bottom left be \( y \), the angle at the top be \( x \), and the angle at the bottom right be \( 83^\circ \). The side opposite \( 83^\circ \) is 14, the side opposite \( x \) is 13, and the base is 13? No, the base is 13, the side with length 14 is one of the equal - ish sides. Wait, maybe the Law of Cosines is better here. Wait, the triangle has sides 13, 14, and the included angle? No, the angle of \( 83^\circ \) is between the side of length 13 and the other side. Wait, let's use Law of Cosines to find the third side first. Let the sides be \( a = 13 \), \( b = 14 \), angle \( C = 83^\circ \) between them. Then the third side \( c \) is given by \( c^{2}=a^{2}+b^{2}-2ab\cos C \)
\(c^{2}=13^{2}+14^{2}-2\times13\times14\times\cos 83^\circ\)
\(c^{2}=169 + 196-364\times0.1219\)
\(c^{2}=365-44.37\)
\(c^{2}=320.63\), \(c\approx17.91\)
Now, use Law of Sines on angle \( x \) (opposite side 13):
\(\frac{\sin x}{13}=\frac{\sin 83^\circ}{17.91}\)
\(\sin x=\frac{13\times\sin 83^\circ}{17.91}\)
\(\sin 83^\circ\approx0.9925\)
\(13\times0.9925 = 12.9025\)
\(\sin x=\frac{12.9025}{17.91}\approx0.7204\)
\(x=\arcsin(0.7204)\approx46.13^\circ\)
Wait, I think my initial side - angle correspondence was wrong. Let's start over. Let's assume: in the triangle, the side of length 13 is opposite the angle \( x \), the side of length 14 is opposite the angle of \( 83^\circ \). Then by Law of Sines:
\(\frac{\sin x}{13}=\frac{\sin 83^\circ}{14}\)
\(\sin x=\frac{13\sin 83^\circ}{14}\)
\(\sin 83^\circ\approx0.992546\)
\(\sin x=\frac{13\times0.992546}{14}=\frac{12.903098}{14}\approx0.92165\)
Wait, that's still more than 0.92, but \(\arcsin(0.92165)\approx67.2^\circ\). But earlier when I calculated the third side, I got a different result. There must be a mis - labeling. Wait, the diagram: the triangle has sides 13 (base), 14 (one of the other sides), and the angle of \( 83^\circ \) is at the base and the side of length 14? No, the angle of \( 83^\circ \) is between the side of length 13 and the side of length (let's say) \( c \), and the side of length 14 is opposite the \( 83^\circ \) angle. So:
Angle \( A = 83^\circ \), side \( a = 14 \), side \( b = 13 \), angle \( B=x \)
Law of Sines: \(\frac{\sin A}{a}=\frac{\sin B}{b}\)
\(\frac{\sin 83^\circ}{14}=\frac{\sin x}{13}\)
\(\sin x=\frac{13\sin 83^\circ}{14}\)
\(\sin 83^\circ\approx0.992546\)
\(\sin x=\frac{13\times0.992546}{14}\approx\frac{12.903098}{14}\approx0.92165\)
\(x=\arcsin(0.92165)\approx67.2^\circ\)
Wait, but the sum of angles in a triangle is \( 180^\circ \). If \( x = 67.2^\circ \) and one angle is \( 83^\circ \), the third angle is \( 180 - 67.2-83=29.8^\circ \), and the side opposite \( 29.8^\circ \) would be calculated by Law of Sines: \(\frac{\sin 29.8^\circ}{c}=\frac{\sin 83^\circ}{14}\), \( c=\frac{14\sin 29.8^\circ}{\sin 83^\circ}\approx\frac{14\times0.497}{0.9925}\approx7.03\), which doesn't match the side of length 13. So my labeling is wrong.
Wait, the correct labeling: the side of length 13 is adjacent to the \( 83^\circ \) angle, the side of length 14 is opposite the \( 83^\circ \) angle? No, the side of length 13 is opposite the \( x \) angle, the side of length 14 is adjacent to the \( 83^\circ \) angle. I think the problem is that I mis - identified the sides. Let's use the Law of Sines correctly. Let's suppose:

Let the triangle have:

- Angle \( C = 83^\circ \)

- Side \( c = 13 \)

- Side \( b = 14 \)

- Angle \( B=x \)

Then by Law of Sines: \(\frac{\sin B}{b}=\frac{\sin C}{c}\)

\(\frac{\sin x}{14}=\frac{\sin 83^\circ}{13}\)

\(\sin x=\frac{14\sin 83^\circ}{13}\)

\(\sin 83^\circ\approx0.992546\)

\(\sin x=\frac{14\times0.992546}{13}=\frac{13.895644}{13}\approx1.0689\)

Ah! Here's the mistake. \(\sin x\) can't be greater than 1. So this means that the side of length 14 cannot be opposite an angle in a triangle with side 13 and angle \( 83^\circ \) opposite side 13. So the correct correspondence is that the side of length 13 is opposite the \( 83^\circ \) angle, and the side of length 14 is opposite the \( x \) angle. Wait, that would mean:

Angle \( A = 83^\circ \), side \( a = 13 \)

Angle \( B=x \), side \( b = 14 \)

But since \( b > a \), then \( B > A \), so \( x>83^\circ \), but \(\sin x=\frac{14\sin 83^\circ}{13}\approx1.0689\) which is impossible. So there must be a mis - interpretation of the diagram. Wait, maybe the angle of \( 83^\circ \) is not opposite the side of length 13. Maybe the triangle is labeled as: the side of length 13 is between the angle \( x \) and the \( 83^\circ \) angle, and the side of length 14 is opposite the \( x \) angle. Wait, I think the user made a typo or the diagram is mis - labeled. Wait, no, let's check again.

Wait, maybe the triangle has sides: 13, 14, and the angle of \( 83^\circ \) is opposite the side of length 14. So:

Angle \( A = 83^\circ \), side \( a = 14 \)

Side \( b = 13 \), angle \( B=x \)

Law of Sines: \(\frac{\sin x}{13}=\frac{\sin 83^\circ}{14}\)

\(\sin x=\frac{13\sin 83^\circ}{14}\approx\frac{13\times0.9925}{14}\approx\frac{12.9025}{14}\approx0.9216\)

\(x=\arcsin(0.9216)\approx67.2^\circ\)

Even though the sum of angles would be \( 67.2 + 83+\text{third angle}=180\), third angle \( = 29.8^\circ \), and the side opposite \( 29.8^\circ \) is \( c \), \(\frac{\sin 29.8^\circ}{c}=\frac{\sin 83^\circ}{14}\), \( c=\frac{14\sin 29.8^\circ}{\sin 83^\circ}\approx\frac{14\times0.497}{0.9925}\approx7.03\), which is less than 13, so \( b = 13 \) is greater than \( c\approx7.03 \), so \( B > C \), which is consistent with \( x = 67.2^\circ>29.8^\circ \). So maybe this is the correct approach.

So, proceeding with \(\sin x=\frac{13\sin 83^\circ}{14}\approx0.9216\)

\(x=\arcsin(0.9216)\approx67.2^\circ\) (rounded to the nearest hundredth)

Wait, but let's calculate it more accurately.

\(\sin 83^\circ=\sin(90 - 7)^\circ=\cos 7^\circ\approx0.9925461516\)

\(13\times0.9925461516 = 12.90309997\)

\(12.90309997\div14\approx0.92165\)

\(\arcsin(0.92165)\): using a calculator, \(\arcsin(0.92165)\approx67.2^\circ\) (since \(\sin 67^\circ\approx0.9205\), \(\sin 67.2^\circ=\sin(67 + 0.2)^\circ=\sin 67^\circ\cos 0.2^\circ+\cos 67^\circ\sin 0.2^\circ\approx0.9205\times0.99997 + 0.3907\times0.00349\approx0.9205+0.00137\approx0.9219\), which is close to 0.92165, so \( x\approx67.2^\circ \))

Answer:

\(x\approx67.20^\circ\) (If we consider the correct side - angle correspondence, the answer is approximately \( 67.20^\circ \))

Wait, but earlier when we thought \( b > a \) leads to impossible sine, but actually, I had the sides reversed. The correct Law of Sines application is: if side \( b = 14 \) is opposite angle \( x \), and side \( a = 13 \) is opposite angle \( 83^\circ \), but since \( 14>13 \), angle \( x>83^\circ \), but \(\sin x=\frac{14\sin 83^\circ}{13}\approx1.0689\) is impossible. So the only way is that the side of length 13 is opposite angle \( x \), and side of length 14 is opposite angle \( 83^\circ \), so \( x<83^\circ \), and \(\sin x=\frac{13\sin 83^\circ}{14}\approx0.9216\), so \( x=\arcsin(0.9216)\approx67.2^\circ\)

So the final answer is approximately \( 67.20^\circ \)