SOLUTION: On a day with no wind, a hot-air balloon hovers at a point above a long, straight river. On the west side of the balloon, a sailboat is spotted in the river at an angle of depressi
Algebra ->
Customizable Word Problem Solvers
-> Geometry
-> SOLUTION: On a day with no wind, a hot-air balloon hovers at a point above a long, straight river. On the west side of the balloon, a sailboat is spotted in the river at an angle of depressi
Log On
Question 1058536: On a day with no wind, a hot-air balloon hovers at a point above a long, straight river. On the west side of the balloon, a sailboat is spotted in the river at an angle of depression of 48 degrees. On the east side, a canoe spots the balloon at an angle of inclination of 29 degrees. The distance between the balloon and the canoe is 650m.
a) What is the height of the balloon?
b) What is the distance between the balloon and the sailboat?
c) What is the distance between the sailboat and the canoe?
My solutions:
Divide triangle BSC into two right angle triangles by drawing a line between angle B (the top) down to the base (line SC) and calling it h and the point where it meets SC as point D. Angle C
is 29 degrees, length BC is 650m, angle S is 48 degrees. Length BS is c.
a) Solving for h (height) sine theta in triangle BCD:
sin(29) = opposite/hypotenuse
sin(29) = h/650
h = 650sin(29)
h = 315.1 meters
b) sin(48) = opposite/hypotenuse in triangle BDS
sin(48) = 315.1/c
c = 315.1/sin(48)
c = 424 meters
c) finding length of SD
tan(48) = 315.1/SD
SD = 315.1/tan(48)
SD = 283.7
finding length DC
tan(29) = 315.1/DC
DC = 315.1/tan(29)
DC = 568.5
283.7 + 568.5 = distance between sailboat and canoe is 852.2 meters.
