Question

Two soccer players on the field are 10 m apart, and their coach is on the sideline. The angle measured from player 1 between player 2 and the coach is 52°. The angle measured from player 2 between player 1 and the coach is 40°. How far is player 1 from the coach?a) 6.4 m b) 7.9 m c) 8.2 m d) 12.3 m

Answer 1

Answer:

The correct option is a) 6.4 m

Step-by-step explanation:

Let A represents the position of first player B represents the position of second player and C represents the position of the Coach,

According to the question,

AB = 10 meters,

$m\angle BAC = 52^{\circ}$

$m\angle CBA = 40^{\circ}$

Since, the sum of all interior angles of a triangle is 180°,

$\implies m\angle CAB+m\angle ABC+m\angle ACB=180^{\circ}$

$52^{\circ}+40^{\circ}+m\angle ACB=180^{\circ}$

$92^{\circ}+m\angle ACB=180^{\circ}$

$m\angle ACB=180^{\circ}-92^{\circ}=88^{\circ}$

Using law of sine,

$\frac{\sin C}{AB}=\frac{\sin B}{AC}$

$\frac{\sin 88^{\circ}}{10}=\frac{\sin 40^{\circ}}{AC}$

$\implies AC = \frac{10\sin 40^{\circ}}{\sin 88^{\circ}}=6.43179\approx 6.4$

Hence, player 1 is 6.4 meters far from the coach.