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

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116