Question

two teams are pulling a heavy chest, located at point x. 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 110°. law of sines: \\(\frac{\sin(a)}{a} = \frac{\sin(b)}{b} = \frac{\sin(c)}{c}\\) which equation can be used to solve for angle a? \\(\bigcirc\\) \\(\frac{\sin (a)}{2.4} = \frac{\sin (110^\circ)}{4.6}\\) \\(\bigcirc\\) \\(\frac{\sin (a)}{4.6} = \frac{\sin (110^\circ)}{2.4}\\) \\(\bigcirc\\) \\(\frac{\sin (a)}{3.2} = \frac{\sin (110^\circ)}{4.6}\\) \\(\bigcirc\\) \\(\frac{\sin (a)}{4.6} = \frac{\sin (110^\circ)}{3.2}\\)

Explanation:

Step1: Match angles to opposite sides

In triangle AXB:

• Angle at X: $110^\circ$, opposite side $AB = 4.6$
• Angle A, opposite side $XB = 3.2$

Step2: Apply Law of Sines

Set up ratio for angle A and $110^\circ$:
$\frac{\sin(A)}{\text{opposite side to } A} = \frac{\sin(110^\circ)}{\text{opposite side to } 110^\circ}$
$\frac{\sin(A)}{3.2} = \frac{\sin(110^\circ)}{4.6}$

Answer:

$\boldsymbol{\frac{\sin(A)}{3.2} = \frac{\sin(110^\circ)}{4.6}}$ (the third option)