Question

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 )

Answer 1

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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Answer 2

$\huge\underbrace\mathfrak \red{ANSWER }$

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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