Question

the angles of elevation of the top of the tower from two points at distances of 6cm and 13.5 cm from the base of the tower and in the same straight line with it , are complementary . find the height of the tower.

Answer 1

Given: AB is the tower.

P and Q are the points at distance of 4m and 9m respectively.

From fig, PB = 4m, QB = 9m.

Let the angle of elevation from P be α and angle of elevation from Q be β.

So, Given that α and β are supplementary. 
Thus, α + β = 90

In triangle ABP,
tan α = AB/BP – (i)

In triangle ABQ,
tan β = AB/BQ

⇒ tan (90 – α) = AB/BQ  (Since, α + β = 90)
⇒ cot α = AB/BQ
⇒ 1/tan α = AB/BQ

So, tan α = BQ/AB – (ii)

Now, From (i) and (ii)

We get, AB/BP = BQ/AB

⇒ AB² = BQ x BP
⇒ AB² = 4 x 9
⇒ AB² = 36

Therefore, AB = 6.

∴ Height of tower is 6m.

There's an alternate method too:

Let the height of the tower be AB=h m.
And, the distances from it's foot be BD = 6 m and BC = 13.5 m respectively.

Also, let ∠ACB = θ and ∠ADB = (90°-θ)  (Since angles are complementary).

Now, In ΔABD

⇒ tan (90° - θ) = h/6
⇒ cotθ = h/6    (Since tan (90° - θ) = cotθ) ----------(1)

And, In ΔABC

⇒ cotθ = 13.5/h ----------(2)

Now, From (1) and (2),

We get, h/6 =13.5/h
⇒h² = 13.5 x 6
∴ h = √81     
      =9 m.

Thus, Height of the tower is 9 meter.

Hope This Helps :)

Answer 2

Given AB is the tower.
P and Q are the points at distance of 4m and 9m respectively.
From fig, PB = 4m, QB = 9m.
Let angle of elevation from P be α and angle of elevation from Q be β.
Given that α and β are supplementary. Thus, α + β = 90
In triangle ABP,
tan α = AB/BP – (i)
In triangle ABQ,
tan β = AB/BQ
tan (90 – α) = AB/BQ (Since, α + β = 90)
cot α = AB/BQ
1/tan α = AB/BQ
So, tan α = BQ/AB – (ii)
From (i) and (ii)
AB/BP = BQ/AB
AB^2 = BQ x BP
AB^2 = 4 x 9
AB^2 = 36
Therefore, AB = 6.
Hence, height of tower is 6m.