The angular elevation of tower CD at a point A due south of it is 60^ o and at a point B due west of A, the elevation is 3 0 ^ o . If AB = 5km, then the height of the tower is (where, C, B and A are on the same ground

Correct answer: D. 5 3 2 2 km From Δ CDA, x = hcot 60 ^ o = (h)/(3) From Δ CDB, y = hcot 30 ^ o = 3 h x^ 2 + 5 ^ 2 = y^ 2 (in ∆ ACB ) ⇒ ((h)/(3))^ 2 + 5 ^ 2 = ( 3 h)^ 2 ⇒ h = 5 3 2 2 km.

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Heights and Distances

NTA Abhyas 2020 · Mathematics · Heights and Distances

The angular elevation of tower CD at a point A due south of it is...

The angular elevation of tower

CD

at a point

A

due south of it is

{\text{60}^{o}

and at a point

B

due west of

A,

the elevation is

3{0}^{o}.

AB = 5km,

then the height of the tower is (where,

C, B

and

A

are on the same ground level) The angular elevation of tower CD at a point A due south of it is 60^ o and at a point B due west of A, the elevation

Options

A $2\sqrt{3}km$
B $2\sqrt{6}km$
C $\dfrac{5\sqrt{3}}{2}km$
D $\dfrac{5\sqrt{3}}{2\sqrt{2}}km$

Correct answer

D. $\dfrac{5\sqrt{3}}{2\sqrt{2}}km$

Step-by-step solution

From

\text{\Delta CDA}, \text{x} = \text{h}cot{60}^{o} = \dfrac{h}{3}

From

\text{\Delta CDB}, \text{y} = \text{h}cot{30}^{o} = \sqrt{3}\text{h}

x^{2} + {5}^{2} = y^{2}

(in

∆ ACB

)

⇒ \left(\dfrac{h}{3}\right)^{2} + {5}^{2} = \left(\sqrt{3}h\right)^{2} ⇒ h = \dfrac{5\sqrt{3}}{2\sqrt{2}}km.

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