A pole of height AB is placed vertically on the hemispherical dome of diameter CD mounted on the ground. CD is produced to the points E and F such that DE = 6 m and EF = 4 m. The angle of elevation of top of the pole from a point E is 45°. The angle of depression of the point F from the bottom of the pole is 30°. Find the height of the pole AB and the height OB of the hemispherical dome . ( √3 = 1.732 )
Answers
Given : A pole of height AB is placed vertic ally on the hemispherical dome of diameter CD mounted on
the ground. CD is produced to the points E and F such that DE = 6 m and EF = 4 m.
The angle of elevation of top of the pole from a point E is 45°. The angle of depression of the point F from the bottom of the pole is 30°
To Find : height of the pole AB and the height OB of the hemispherical dome
Solution:
Assuming pole is at middle of dome.
Let say Radius of Dome = R
=> OB = OC = OD = R
OA = OB + AB = R + AB
OE = OD + DE = R + 6 m
OF = OD + DE + EF = R + 6 + 4 = R + 10 m
Tan 45 = OA/OE
=> 1 = OA/OE
=>OA = OE
=> R + AB = R + 6
=> AB = 6
Tan 30 = OB/OF
=> 1/√3 = R/ R + 10
=> R + 10 = R√3
=> R (√3 - 1) = 10
=> R = 10/ (√3 - 1)
=> R = 10 (√3+ 1) / 2
=> R = 5 (√3+ 1)
=> R = 5(1.732 + 1)
=> R = 5(2.732)
=> R = 13.66
= OB = 13.66
Assuming bottom of pole make tangent
Let say Radius of Dome = R
=> OB = OC = OD = R
OE = OD + DE = R + 6 m
OF = OD + DE + EF = R + 6 + 4 = R + 10 m
BM ⊥ CD
The angle of depression of the point F from the bottom of the pole is 30°.
Sin 30 = OB/OF
=> 1/2 = R / R + 10
=> R + 10 = 2R
=> R = 10
OB = 10
OF = 20
BE = 10√3 (BE² = OB² - OE²)
BM ⊥ CD
=> BM = 10 x 10√3/20 = 5√3
Tan 60 = BM/OM
=> √3 = 5√3/OM
=> OM = 5
Tan 45 = MA/ME
MA = AB + BM = AB + 5√3
ME = OE - OM = R + 6 - 5 = 10 + 6 -5 = 11
=> 1 = (AB + 5√3)/11
=>AB = 11 - 5√3
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Given : A pole of height AB is placed vertic ally on the hemispherical dome of diameter CD mounted on
the ground. CD is produced to the points E and F such that DE = 6 m and EF = 4 m.
The angle of elevation of top of the pole from a point E is 45°. The angle of depression of the point F from the bottom of the pole is 30°
To Find : height of the pole AB and the height OB of the hemispherical dome
Solution:
Assuming pole is at middle of dome.
Let say Radius of Dome = R
=> OB = OC = OD = R
OA = OB + AB = R + AB
OE = OD + DE = R + 6 m
OF = OD + DE + EF = R + 6 + 4 = R + 10 m
Tan 45 = OA/OE
=> 1 = OA/OE
=>OA = OE
=> R + AB = R + 6
=> AB = 6
Tan 30 = OB/OF
=> 1/√3 = R/ R + 10
=> R + 10 = R√3
=> R (√3 - 1) = 10
=> R = 10/ (√3 - 1)
=> R = 10 (√3+ 1) / 2
=> R = 5 (√3+ 1)
=> R = 5(1.732 + 1)
=> R = 5(2.732)
=> R = 13.66
= OB = 13.66
Assuming bottom of pole make tangent
Let say Radius of Dome = R
=> OB = OC = OD = R
OE = OD + DE = R + 6 m
OF = OD + DE + EF = R + 6 + 4 = R + 10 m
BM ⊥ CD
The angle of depression of the point F from the bottom of the pole is 30°.
Sin 30 = OB/OF
=> 1/2 = R / R + 10
=> R + 10 = 2R
=> R = 10
OB = 10
OF = 20
BE = 10√3 (BE² = OB² - OE²)
BM ⊥ CD
=> BM = 10 x 10√3/20 = 5√3
Tan 60 = BM/OM
=> √3 = 5√3/OM
=> OM = 5
Tan 45 = MA/ME
MA = AB + BM = AB + 5√3
ME = OE - OM = R + 6 - 5 = 10 + 6 -5 = 11
=> 1 = (AB + 5√3)/11
=>AB = 11 - 5√3
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