what is the height of the building if the 1st angle of elevation is 50 degrees & the 2nd is 28 degrees were taken at distance 100 feet apart?
回答 (3)
2019-03-13 9:33 am
Refer to the diagram below.
In ΔABC:
tan28° = h/(x + 100)
h/tan28° = x + 100 …… [1]
In ΔDBC:
tan50° = h/x
h/tan50° = x …… [2]
[1] - [2]:
(h/tan28°) - (h/tan50°) = 100
h [(1/tan28°) - (1/tan50°)] = 100
h = 100 / [(1/tan28°) - (1/tan50°)]
h = 96
Height of the building (BC) = 96 ft
In ΔABC:
tan28° = h/(x + 100)
h/tan28° = x + 100 …… [1]
In ΔDBC:
tan50° = h/x
h/tan50° = x …… [2]
[1] - [2]:
(h/tan28°) - (h/tan50°) = 100
h [(1/tan28°) - (1/tan50°)] = 100
h = 100 / [(1/tan28°) - (1/tan50°)]
h = 96
Height of the building (BC) = 96 ft
2019-03-14 2:59 am
tan 50⁰ = y/x
tan 28⁰ = y/(x + 100)
1•19 x = y
0•53 x + 53 = y
0•66x = 53
x = 80•3 ft
y = x tan 50⁰
y = 95•7 ft______height of building.
tan 28⁰ = y/(x + 100)
1•19 x = y
0•53 x + 53 = y
0•66x = 53
x = 80•3 ft
y = x tan 50⁰
y = 95•7 ft______height of building.
2019-03-13 12:41 pm
What is the height of the building if the 1st angle of elevation is 50 degrees &
the 2nd is 28 degrees were taken at distance 100 feet apart?
tan 50 = h/x
tan 28 = h/(x + 100)
x tan 50 = (x + 100) tan 28
Solution:
x ≈ 80.557
h = x tan 50 = 96 ft
the 2nd is 28 degrees were taken at distance 100 feet apart?
tan 50 = h/x
tan 28 = h/(x + 100)
x tan 50 = (x + 100) tan 28
Solution:
x ≈ 80.557
h = x tan 50 = 96 ft
收錄日期: 2021-05-01 22:30:22
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