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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].

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

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

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

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


Sagot :

To solve the problem using the Law of Sines, let's first identify the given distances in the context of a triangle formed by the points where the ropes are attached and the chest.

1. The distance between Team A and Team B ([tex]\(AB\)[/tex]) is 4.6 meters.
2. The distance from Team A to the chest ([tex]\(AX\)[/tex]) is 2.4 meters.
3. The distance from Team B to the chest ([tex]\(BX\)[/tex]) is 3.2 meters.
4. The angle between the teams' ropes is [tex]\(110^\circ\)[/tex].

We will use the Law of Sines which states:
[tex]\[ \frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}. \][/tex]
In our case, angle [tex]\(C\)[/tex] is [tex]\(110^\circ\)[/tex]. We need to set up an equation to find angle [tex]\(A\)[/tex], which is the angle opposite to side [tex]\(AX\)[/tex] (2.4 meters).

To use the Law of Sines:
[tex]\[ \frac{\sin(A)}{AX} = \frac{\sin(C)}{AB} \][/tex]
Substitute the known values into the equation:
[tex]\[ \frac{\sin(A)}{2.4} = \frac{\sin(110^\circ)}{4.6}. \][/tex]

Therefore, the correct equation to solve for angle [tex]\(A\)[/tex] using the Law of Sines is:
[tex]\[ \frac{\sin(A)}{2.4}=\frac{\sin \left(110^{\circ}\right)}{4.6}. \][/tex]
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