Question

The angles of elevation of a top of a tower from two points at a distance of a 'a' meter and 'b' meters from the base of the tower along the same straight line are complementary . Prove that the height of the tower is √ab.


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Answer 1

Step-by-step explanation:

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Answer 2

Answer- The above question is from the chapter 'Some Applications of Trigonometry'.

Given question: The angles of elevation of a top of a tower from two points at a distance of a 'a' meter and 'b' meters from the base of the tower along the same straight line are complementary. Prove that the height of the tower is √ab.

Solution: (Diagram has been attached.)

Let AB = Height of tower

BC = Distance between foot of tower and point 1 = a m

BD = Distance between foot of tower and point 2 = b m

CD = Distance between two points = (a - b) m

Let ∠ACB = θ

⇒ ∠ADB = 90° - θ (∵ the two angles are complementary.)

To prove: AB = √ab m

Sol.: In ΔABC, $\dfrac{AB}{BC} = tan \: \theta$

⇒ AB = BC tan θ

⇒ AB = a tan θ --- (1)

In ΔABD, $\dfrac{AB}{BD}  = tan \:  (90 ^ {\circ} - \theta )$

⇒ AB = BD tan (90° - θ)

⇒ AB = b cot θ --- (2) [∵ tan (90° - θ) = cot θ]

Equation 1 × Equation 2,

AB² = a tan θ × b cot θ

AB² = ab × 1 [∵ tan θ = $\frac{1}{cot \theta}$ ]

AB = √ab m

∴ Height of the tower = √ab m.