Two fire-lookout stations are 190 miles apart, with station a directly south of station b. both stations spot a fire. the bearing of the fire from station a is upper n 55 degrees upper e and the bearing of the fire from station b is upper s 60 degrees e. how far, to the nearest tenth of a mile, is the fire from each lookout station?

Step-by-step explanation:

Let's say the position of the fire is point C.

Bearings are measured from the north-south line.  So ∠BAC = 55°, and ∠ABC = 60°.

Since angles of a triangle add up to 180°, ∠ACB = 65°.

Using law of sine:

190 / sin 65° = a / sin 60° = b / sin 55°

Solving:

a = 181.6

b = 171.7

Station A is 181.6 miles from the fire and station B is 171.7 miles from the fire.

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."