Two fire-lookout stations are 18 miles apart, with station B directly east of station A. Both stations spot a fire. The bearing of the fire from station A is N35°E

Question

and the bearing of the fire from station B is N30°W.

How far, to the nearest tenth of a mile, is the fire from each lookout station?

Answer 1

Patrick has the right idea, but since B is east of A, angle B is 90° - 30° = 60°, and angle A is 90° - 35° = 55° (he also made a slight typo with his method, as 180 - 35 - 30 = 115)

Then, letting the fire be at c, angle C is 180° - 60° - 55° = 65°

We can then follow Patrick's process.

c = AB = 18

b = AC = 18 sin B/sin C = 18 sin 60°/sin 65° = 17.1999595403339

a = BC = 18 sin A/sin C = 18 sin 55°/sin 65° = 16.2690170029463

The distance from A to the fire is 17.1999595403339

The distance from B to the fire is 16.2690170029463

Answer 2

Law of sines

sin 30 / x = sin 125 / 18

sin 30 * 18 = sin 125 * x

sin 30 * 18/ sin 125 = x

x is ALMOST 11 miles

sin 35/y = sin 125 / 18

sin 35 * 18 / sin 125 = y

y is slightly more than 12.6 miles