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1. A point on the ceiling of a room is 12ft. from each of the vertices of a triangle on the floor whose legs are 6ft. A and 8ft...Asked by Anonymous
A point on the ceiling of a room is 12ft. from each of the vertices of a triangle on the floor whose legs are 6ft. A and 8ft. respectively. find the length of the ceiling.
2. With a 12ft. pole marked in feet, how can one determine the foot of the perpendicular let fall to the floor from a point on a ceiling of a room 9ft. high?
3. A hill slopes down from a building with a grade of one for to five feet measured along the horizontal (slope of 1/5). If a ladder 36 ft. long is set against the building, with its foot 12ft. down the hill. How high will it reach the building?
2. With a 12ft. pole marked in feet, how can one determine the foot of the perpendicular let fall to the floor from a point on a ceiling of a room 9ft. high?
3. A hill slopes down from a building with a grade of one for to five feet measured along the horizontal (slope of 1/5). If a ladder 36 ft. long is set against the building, with its foot 12ft. down the hill. How high will it reach the building?
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Answered by
Steve
#1. How can working with a single point on the ceiling tell anything about the length of the ceiling?
#2. Drop the ladder to the floor, and mark where it hits. The use the ladder to construct a hexagon. Two of the main diagonals intersect at the center of the hexagon, just below the point on the ceiling.
#3. Draw a diagram. Let the ladder touch the hill at point P. Label the top of the building T. Drop a vertical from T, which intersects the hill at H, and extends to point Q, where PQ is horizontal.
If tan(x) = 1/5, then
PQ = 12 cos(x)
HQ = 12 sin(x)
TQ = √(36^2-PQ^2)
TH = TQ-HQ = height of building
#2. Drop the ladder to the floor, and mark where it hits. The use the ladder to construct a hexagon. Two of the main diagonals intersect at the center of the hexagon, just below the point on the ceiling.
#3. Draw a diagram. Let the ladder touch the hill at point P. Label the top of the building T. Drop a vertical from T, which intersects the hill at H, and extends to point Q, where PQ is horizontal.
If tan(x) = 1/5, then
PQ = 12 cos(x)
HQ = 12 sin(x)
TQ = √(36^2-PQ^2)
TH = TQ-HQ = height of building
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