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°. which equation can be used to solve for angle a? law of sines: $\frac{sin(a)}{a}=\frac{sin(b)}{b}=\frac{sin(c)}{c}$

Explanation:

Step1: Recall the Law of Sines

The Law of Sines states that in a triangle, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$, where $A$, $B$, $C$ are angles and $a$, $b$, $c$ are the lengths of the sides opposite to those angles respectively.

Step2: Identify sides and angles in the given triangle

In the triangle formed by the two - teams and the chest, let the angle we want to find be $A$, the side opposite to $A$ have length $a = 3.2$, the known angle be $B=110^{\circ}$ and the side opposite to $B$ have length $b = 4.6$.

Step3: Apply the Law of Sines

According to the Law of Sines, $\frac{\sin A}{3.2}=\frac{\sin(110^{\circ})}{4.6}$, which can be rewritten as $\frac{\sin(A)}{3.2}=\frac{\sin(110^{\circ})}{4.6}$ or $\frac{\sin(A)}{3.2}\times3.2=\frac{\sin(110^{\circ})}{4.6}\times3.2$.

Answer:

$\frac{\sin(A)}{3.2}=\frac{\sin(110^{\circ})}{4.6}$ (corresponding to the fourth option in the original multiple - choice list, assuming the order of options as presented in the image)