Two teams are pulling a heavy chest, located at point [tex]\(X\)[/tex]. The teams are 4.6 meters away from each other. Team A is 2.4 meters away from the chest, and Team B 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 this problem, let's use the Law of Sines properly applied to the given scenario. The Law of Sines states:

[tex]\[ \frac{\sin (A)}{a}=\frac{\sin (B)}{b}=\frac{\sin (C)}{c} \][/tex]

Given the problem:

- Team A is 2.4 meters away from the chest.
- Team B is 3.2 meters away from the chest.
- The teams are 4.6 meters away from each other.
- The angle between their ropes (angle [tex]\(C\)[/tex]) is [tex]\(110^{\circ}\)[/tex].

Here [tex]\(a, b, c\)[/tex] are the sides of the triangle, and [tex]\(A, B, C\)[/tex] are the angles opposite those sides respectively. We are given:

- [tex]\(a = 2.4 \text{ meters}\)[/tex] (distance from Team A to the chest)
- [tex]\(b = 3.2 \text{ meters}\)[/tex] (distance from Team B to the chest)
- [tex]\(c = 4.6 \text{ meters}\)[/tex] (distance between the teams)

Also, we know [tex]\( \angle C = 110^{\circ} \)[/tex].

We are looking for the equation to solve for angle [tex]\(A\)[/tex]. According to the law of sines, we have:

[tex]\[ \frac{\sin (A)}{a}=\frac{\sin (C)}{c} \][/tex]

Plugging in the known values:

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

Thus, the correct equation that can be used to solve for angle [tex]\(A\)[/tex] is:

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

Therefore, the correct answer is:

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