The upper(3/4) th portion of a vertical pole subtends an angle tan^(-1)(3/5) at a point in the horizontal plane through its foot

Question

The upper \(\left(\frac{3}{4}\right)\)th portion of a vertical pole subtends an angle tan^-1 \(\left(\frac{3}{5}\right)\) at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is

(a) 20 m 

(b) 40 m 

(c) 60 m 

(d) 80 m

Answer 1

(b) 40 m

Let PQ be the vertical pole of height h metres. 

Let A be a point on PQ such that PA = \(\frac{3}{4}h\), 

and AQ = \(h-\frac{3}{4}h\) = \(\frac{1}{4}h\).

Let the portion PA of the pole subtend an angle θ₁ at the observation point O such that QO = 40 m.

Let AQ subtend θ₂ at 0.

Then,

θ₁ = tan^-1\((\frac{3}{5})\) 

⇒  tan θ₁ = \(\frac{3}{5}\) ...(i)

In rt. Δ AOQ,

tan θ₂ = \(\frac{AQ}{QO}\) 

⇒ tan θ₂ = \(\frac{h}{160}\) ...(ii)

In rt. Δ POQ,

tan (θ₁ + θ₂) = \(\frac{PQ}{QO}=\frac{h}{40}\)

⇒ \(\frac{tan\,\theta_1\,+\,tan\,\theta_2}{1-\,tan\,\theta_1\,tan\,\theta_2}=\frac{h}{40}\) [From eqns (i) and (ii)]

⇒ \(\frac{5(h\,+\,96)}{800\,-\,3h}=\) \(\frac{h}{40}\)

⇒ h² – 200h + 6400 = 0

⇒ (h – 160) (h – 40) = 0

⇒ h = 160 or h = 40.

According to the given options one of the possible heights = 40 m.