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:

Answer 1

The correct answer is:

D. $\frac{sin(A)}{4.6}=\frac{sin(110)}{3.2}$

$|Huntrw6|$

Answer 2

For a better understanding of the question, please see attached picture that I've created.
We need to identify the other two angles by applying the Law of sine where we are given with the following values:
a = 3.2 unit
b = 2.4 unit
c = 4.6 unit
∠A = unknown
∠B = unknown
∠C = 110°

Solving for ∠B, we have:
sin C / c = sin B / b
sin 110 / 4.6 = sin B / 2.4
sin B = 2.4*sin 110 / 4.6
B = 29.36°

Solving for ∠A, we have:
sinC/c = sinA/ a
sin110 / 4.6 = sinA / 3.2
sinA = 3.2*sin110/ 4.6
A = 40.82°

Therefore, the missing angles are ∠A=40.82° and ∠B=29.36°.