Suppose you need to find the height of a tall building. Standing 15 meters away from the base of the building, you aim a laser pointer at the closest part of the top of the building. You measure that the laser pointer is 7◦
tilted from pointing straight up. The laser pointer is held 2 meters above the ground. How tall is the building?
PreCalc Need Help?
2020-08-01 11:42 pm
回答 (3)
2020-08-02 12:47 am
Refer to the figure below.
Point A is the laser pointer, and point B is the closest part of the top of the building.
In ΔABC:
tan(∠BAC) = BC/AC
tan(90° - 7°) = BC/(15 m)
BC = 15 tan(83°) m
BC = 122 m
Height of the building
= (122 + 2) m
= 124 m
Point A is the laser pointer, and point B is the closest part of the top of the building.
In ΔABC:
tan(∠BAC) = BC/AC
tan(90° - 7°) = BC/(15 m)
BC = 15 tan(83°) m
BC = 122 m
Height of the building
= (122 + 2) m
= 124 m
2020-08-02 12:04 am
SOHCAHTOA so you want Tan
Tan 83 = Opp/15; Opp = 15 Tan 83 = 122.1651
Then add 2 m = 124 m
Tan 83 = Opp/15; Opp = 15 Tan 83 = 122.1651
Then add 2 m = 124 m
2020-08-02 12:03 am
tan 7 = 15 / h
h = 15 / tan 7 = 122 m
add 2 to get 124 m
h = 15 / tan 7 = 122 m
add 2 to get 124 m
收錄日期: 2021-04-24 07:54:26
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