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question 6
question 6
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started by
Jyoti Pakianathan
on 24 Jan 13
Jyoti Pakianathan
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#1
Jyoti Pakianathan
on 24 Jan 13
If the height of a pole is 2√3 metres and the length of its shadow is 2 metres,
find the angle of elevation of the sun.
If the height of a pole is 2√3 metres and the length of its shadow is 2 metres,
find the angle of elevation of the sun.
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#2
David Kim
on 25 Jan 13
tan(x) = 2√3/2
x = tan-1(2√3/2)
Angle of Elevation = 60
tan(x) = 2√3/2
x = tan-1(2√3/2)
Angle of Elevation = 60
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#3
kwonteddy17
on 25 Jan 13
tan-1(2√3/2) = x
x=60
tan-1(2√3/2) = x
x=60
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#4
Elaine Kim
on 25 Jan 13
tan-1 (2√3/2) = x
x = 60
Elevation of sun = 60
tan-1 (2√3/2) = x
x = 60
Elevation of sun = 60
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#5
Haley Park
on 25 Jan 13
tan x= 2√3/2.
2√3/2.=1.73
tan-1*1.73= 59.97
The elevation of the sun= 59.97
tan x= 2√3/2.
2√3/2.=1.73
tan-1*1.73= 59.97
The elevation of the sun= 59.97
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#6
backchanyel17
on 25 Jan 13
tan-1(2√3/2)= angle of elevation
answer= 60
tan-1(2√3/2)= angle of elevation
answer= 60
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#7
Alex Kim
on 25 Jan 13
Tan-1 (2√3/2) = 60
Tan-1 (2√3/2) = 60
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#8
anonymous
on 25 Jan 13
Tan (x) = 2 root 3 / 2
Tan-1 (2 root 3 / 2) = x
Answer: 60 meters
Tan (x) = 2 root 3 / 2
Tan-1 (2 root 3 / 2) = x
Answer: 60 meters
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#9
Sally Rho
on 25 Jan 13
tan x=2√3/2
2√3/2=1.73
tan-1(2√3/2)=x
x=60
tan x=2√3/2
2√3/2=1.73
tan-1(2√3/2)=x
x=60
...
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#11
Janice Min
on 25 Jan 13
Tan-1 (2√3/2)
angle of elevation = 60
Tan-1 (2√3/2)
angle of elevation = 60
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#12
Harine Choi
on 25 Jan 13
tan x = 2√3/2
x = 60
tan x = 2√3/2
x = 60
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#13
Jason Kim
on 25 Jan 13
tan x = 2√3 over 2
inverse tan 2√3 over 2
x= 60
tan x = 2√3 over 2
inverse tan 2√3 over 2
x= 60
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find the angle of elevation of the sun.
x = tan-1(2√3/2)
Angle of Elevation = 60
x=60
x = 60
Elevation of sun = 60
2√3/2.=1.73
tan-1*1.73= 59.97
The elevation of the sun= 59.97
answer= 60
Tan-1 (2 root 3 / 2) = x
Answer: 60 meters
2√3/2=1.73
tan-1(2√3/2)=x
x=60
angle of elevation = 60
x = 60
inverse tan 2√3 over 2
x= 60