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Two lifeguard towers are 546 yards apart, with tower B directly east of tower A. Both towers spot a hazard in the ocean. The bearing of the hazard from tower A is N35ºE, and the bearing of the hazard from tower B is N49ºW. Which formula gives the distance of the hazard from tower B?
A. [tex]a=546\frac{sin84^{\circ}}{sin55^{\circ}}[/tex]
B. [tex]a=546\frac{sin49^{\circ}}{sin96^{\circ}}[/tex]
C. [tex]a=546\frac{sin35^{\circ}}{sin96^{\circ}}[/tex]
D. [tex]a=546\frac{sin55^{\circ}}{sin84^{\circ}}[/tex]

Answer :

Using the law of sines, the formula that gives the distance of the hazard from tower B is:

C. a = 546sin(35º)/sin(96º).

What is the law of sines?

Suppose we have a triangle in which:

  • The length of the side opposite to angle A is a.
  • The length of the side opposite to angle B is b.
  • The length of the side opposite to angle C is c.

The lengths and the sine of the angles are related as follows:

[tex]\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}[/tex]

From the situation described, we have that:

  • The angles are of 35º, 49º and 180 - (35 + 49) = 96º.
  • The side opposite to the angle of 96º is of 546 yards.
  • The angle opposite to the distance of the hazard to tower B is of 35º.

Hence the relation is:

sin(96º)/546 = sin(35º)/a

a = 546sin(35º)/sin(96º).

Which means that option C is correct.

More can be learned about the law of sines at https://brainly.com/question/25535771

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