Two teams are pulling a heavy chest located at point [tex]$X$[/tex]. The teams are 4.6 meters away from each other. Team [tex]$A$[/tex] is 2.4 meters away from the chest, and Team [tex]$B$[/tex] is 3.2 meters away. Their ropes are attached at an angle of [tex]$110^{\circ}$[/tex].

Law of Sines: [tex]$\frac{\sin (A)}{a}=\frac{\sin (B)}{b}=\frac{\sin (C)}{c}$[/tex]

Which equation can be used to solve for angle [tex]$A$[/tex]?

A. [tex]$\frac{\sin (A)}{2.4}=\frac{\sin \left(110^{\circ}\right)}{4.6}$[/tex]

B. [tex]$\frac{\sin (A)}{4.6}=\frac{\sin \left(110^{\circ}\right)}{2.4}$[/tex]

C. [tex]$\frac{\sin (A)}{3.2}=\frac{\sin \left(110^{\circ}\right)}{4.6}$[/tex]

D. [tex]$\frac{\sin (A)}{4.6}=\frac{\sin \left(110^{\circ}\right)}{3.2}$[/tex]



Answer :

To solve for angle [tex]\( A \)[/tex] using the Law of Sines, we can follow these steps:

1. Identify the given values:
- Distance from Team A to the chest ([tex]\(a\)[/tex]): [tex]\(2.4\)[/tex] meters
- Distance from Team B to the chest ([tex]\(b\)[/tex]): [tex]\(3.2\)[/tex] meters
- Distance between the teams ([tex]\(c\)[/tex]): [tex]\(4.6\)[/tex] meters
- Angle opposite to side [tex]\(c\)[/tex]: [tex]\(\angle C = 110^\circ\)[/tex]

2. Write the Law of Sines formula:
The Law of Sines states that:
[tex]\[ \frac{\sin (A)}{a} = \frac{\sin (B)}{b} = \frac{\sin (C)}{c} \][/tex]

3. Substitute the given values into the Law of Sines:
We need to find an equation that helps us solve for angle [tex]\( A \)[/tex]. Given the side [tex]\(a\)[/tex] and the angle [tex]\(C\)[/tex] opposite to side [tex]\(c\)[/tex], we get:
[tex]\[ \frac{\sin (A)}{2.4} = \frac{\sin (110^\circ)}{4.6} \][/tex]

4. Conclude with the correct equation:
The correct equation to use for solving angle [tex]\( A \)[/tex] is:
[tex]\[ \frac{\sin (A)}{2.4} = \frac{\sin (110^\circ)}{4.6} \][/tex]

Therefore, the equation that can be used to solve for angle [tex]\( A \)[/tex] is:
[tex]\[ \boxed{\frac{\sin (A)}{2.4} = \frac{\sin (110^\circ)}{4.6}} \][/tex]