Question

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. $a=546\frac{sin84^{\circ}}{sin55^{\circ}}$
B. $a=546\frac{sin49^{\circ}}{sin96^{\circ}}$
C. $a=546\frac{sin35^{\circ}}{sin96^{\circ}}$
D. $a=546\frac{sin55^{\circ}}{sin84^{\circ}}$

Answer 1

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:

$\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}$

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

#SPJ1