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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].

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 :

GinnyAnswer
To determine which equation can be used to solve for angle [tex]\( A \)[/tex], let's start by identifying the relevant components in the given problem.

We have:
- Side [tex]\( a = 2.4 \)[/tex] meters
- Side [tex]\( b = 3.2 \)[/tex] meters
- Side [tex]\( c = 4.6 \)[/tex] meters, which is the distance between the two teams
- Angle [tex]\( C = 110^\circ \)[/tex], which is the angle between the ropes where the chest is located

Using the law of sines, we have the following relationship:
[tex]\[ \frac{\sin(A)}{a} = \frac{\sin(C)}{c} \quad \text{or} \quad \frac{\sin(A)}{a} = \frac{\sin(110^\circ)}{4.6} \][/tex]

Given that [tex]\( a = 2.4 \)[/tex] meters, we can substitute [tex]\( a \)[/tex] and [tex]\( C = 110^\circ \)[/tex] into the equation to solve for angle [tex]\( A \)[/tex]:
[tex]\[ \frac{\sin(A)}{2.4} = \frac{\sin(110^\circ)}{4.6} \][/tex]

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

Thus, the answer is:
[tex]\[ \boxed{\frac{\sin (A)}{2.4}=\frac{\sin \left(110^{\circ}\right)}{4.6}} \][/tex]

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