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: Which equation can be used to solve for angle A?

Answer 1

The correct answer is D. or sin110 / 4.6 = sinA / 3.2

Answer 2

Answer: $\frac{sinA}{3.2}=\frac{sin110^{\circ}}{4.6}$

$\text{or }sinA=3.2 \times\frac{sin110^{\circ}}{4.6}$

Step-by-step explanation:

By the sin law,

If a, b and c are the sides of a triangle ABC,

Then by the sine law,

$\frac{sinA}{a} = \frac{sin B}{b}=\frac{sinC}{c}$

According to the question,

AB = 4.6 meter,

AX = 2.4 meter,

BX = 3.2 meter,

m∠AXB = 110°

Hence, by the sine law,

$\frac{sinA}{3.2}=\frac{sin110^{\circ}}{4.6}$

$sinA=3.2 \times\frac{sin110^{\circ}}{4.6}$

Which is the required equation for finding the angle A.