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: $\frac{sin(a)}{a}=\frac{sin(b)}{b}=\frac{sin(c)}{c}$ $\frac{sin(a)}{2.4}=\frac{sin(110^{circ})}{4.6}$ $\frac{sin(a)}{4.6}=\frac{sin(110^{circ})}{2.4}$ $\frac{sin(a)}{3.2}=\frac{sin(110^{circ})}{4.6}$ $\frac{sin(a)}{4.6}=\frac{sin(110^{circ})}{3.2}$

Explanation:

Step1: Recall law of sines

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

Step2: Identify sides and angles

In the given triangle, the side opposite angle $A$ has length $2.4$, and the side opposite the $110^{\circ}$ angle has length $4.6$.

Step3: Write the law - of - sines equation for angle $A$

Using the law of sines, we get $\frac{\sin(A)}{2.4}=\frac{\sin(110^{\circ})}{4.6}$.

Answer:

$\frac{\sin(A)}{2.4}=\frac{\sin(110^{\circ})}{4.6}$ (corresponding to the first option in the multiple - choice list)