Two fire-lookout stations are 14 miles apart, with station B directly east of station A. Both stations spot a fire. The bearing of the fire from station A is Upper N 35 degrees E and the bearing of the fire from station B is N 30 degrees W. How far, to the nearest tenth of a mile, is the fire from each lookout station?

AC = 13.38 miles

BC = 12.65 miles

Where C is the fire location

Step-by-step explanation:

Two fire-lookout stations A and B are 14 miles apart.

The bearing of the fire from station A is Upper N 35 degrees E .

This implies that angle A = 90-35= 55°( reason because B is directly NE of A)

bearing of the fire from station B is N 30 degrees W.

This signifies that angle B is 90-30 =60°

Let's call the fire position C

Angle C = 180-60-55

Angle C= 65°

For Side AC

AB/sin C = AC/sin B

14/sin 65 = AC/sin 60

AC = 13.38 miles

For Side BC

BC/sin A = AB/sin C

BC = sin 55 * (14/sin 65)

BC = 12.65 miles

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation:

y=-2x0 is most likey your asnwer