Posted by Nikki on .
two forest fire stations,P and Q are 20.0 km apart, A ranger at station Q sees a fire 15.0 km away, if the angle between line PQ and the line form P to the fire is 25 , how far to the nearest tenth of a kilometre is statin P from the fire
Not enough information.
Let the fire be at point F.
If the angle at P were 36.9 degrees, then we would have a right triangle with legs 15 and 20, and hypotenuse PF = 25.
However, since angle P is less than 36.9 degrees, a circle of radius 15 will cut the line PF extended in two places, either one of which would be a solution. The angle at F could be either acute or obtuse.
However, using the law of sines,
15/sin15° = 35.5 = 20/sinF
sinF = 20/35.5
F = 34° or 146°
PF/sin(180-(25+34)) = 35.5 makes PF = 30.42
PF/sin(180-(25+146)) = 35.5 makes PF = 5.38
SO, the distance PF depends on which direction the ranger was looking when he saw the fire.