Question

A T. V. Tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60 degree. From a point 20 M away this point on the same bank, the angle of elevation of the top of the tower is 30 degree. Find the height of the tower and the width of the river

Answer 1

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Let height of the tower be = h meters

.·. AB = h

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Now, in ΔABC,

Tan60 = AB/BC = h/BC

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We know that, Tan60 = √3

➟ √3 = h/BC

h = √3BC ----2

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Now, in ΔADB,

Tan30 = AB/DB = h/20 + x

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We know that, Tan30 = 1/√3

➟ 1/√3 = h/20 + x

By cross multiplication,

20 + BC = √3h -----1

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Equating 1 and 2 :-

20 + BC = √3 x √3BC

20 + BC = 3BC

20 = 3BC - BC

20 = 2BC

BC = 20/2

BC = 10

-- Width of river = 10m

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From 1, we know that,

h = √3BC

➟ h = √3 x 10

h = 10√3

-- Height of tower = 10√3m

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Answer 2

• Consider a ∆ABD, having AB = height of T.V. tower.

$\angle{ACB}$ = 60° and $\angle{ADB}$ = 30°

DC = 20 m

• Let BC = x (width of river)

*Refer the attachment for figure.

In ∆ABC

→ tan60° = $\dfrac{AB}{BC}$

→ √3 = $\dfrac{h}{x}$

Cross-multiply them

→ h = x√3 _________ (eq 1)

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In ∆ABD

→ tan30° = $\dfrac{AB}{BD}$

→ $\dfrac{1}{\sqrt{3}}$ = $\dfrac{AB}{BC\:+\:CD}$

→ $\dfrac{1}{\sqrt{3}}$ = $\dfrac{h}{x\:+\:20}$

Cross-multiply them

→ h√3 = x + 20

→ (x√3)√3 = x + 20 [From (eq 1)]

→ 3x = x + 20

→ 3x - x = 20

→ 2x = 20

→ x = 10 m

Put value of x in (eq 1)

→ h = 10√3 m

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Height of tower = 10√3 m and width of river = 10 m

________ [ ANSWER ]

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