A vertical pole subtends an angle tan^(–1)(1/2) at a point P on the ground. If the angles subtended by the upper half and lower half of the ​

Question

A vertical pole subtends an angle tan^–1 \(\left(\frac{1}{2}\right)\) at a point P on the ground. If the angles subtended by the upper half and lower half of the pole at P are respectively α and β, then (tan α, tan β) is equal to

(a) \(\left(\frac{1}{4},\frac{1}{5}\right)\)

(b) \(\left(\frac{1}{5},\frac{2}{9}\right)\)

(c) \(\left(\frac{2}{9},\frac{1}{4}\right)\)

(d) \(\left(\frac{1}{4},\frac{2}{9}\right)\)

Answer 1

(c) \(\left(\frac{2}{9},\frac{1}{4}\right)\)

Let AC be the pole and point P be the position on the ground. Then CB subtends α at P, BA subtends β at P.

Given, 

∠CPA = θ.

Given,

θ = tan^-1 \(\frac{1}{2}\)

⇒ tan θ= \(\frac{1}{2}\)

Also,

θ = α + β

⇒ tan θ = tan (α + β)

⇒ \(\frac{1}{2}\) = \(\frac{tan\,α\,+\,tan\,β}{1-tan\,α\,tan\,β}\)

Checking on all the given options,

(a) When (tan α, tan β) = \(\left(\frac{1}{4},\frac{1}{5}\right)\)

R.H.S = \(\frac{\frac{1}{4}\,+\,\frac{1}{5}}{1-\frac{1}{4}\times\frac{1}{5}}\)

= \(\frac{\frac{9}{20}}{\frac{19}{20}}\)

= \(\frac{9}{19}\) ≠ \(\frac{1}{2}\), not true

(b) When (tan α, tan β) = \(\left(\frac{1}{5},\frac{2}{9}\right)\)

R.H.S = \(\frac{\frac{1}{5}\,+\,\frac{2}{9}}{1-\frac{1}{5}\times\frac{2}{9}}\)

= \(\frac{\frac{19}{45}}{\frac{43}{45}}\)

= \(\frac{19}{43}\) ≠ \(\frac{1}{2}\), not true

(c) When (tan α, tan β) = \(\left(\frac{2}{9},\frac{1}{4}\right)\)

 R.H.S = \(\frac{\frac{2}{9}\,+\,\frac{1}{4}}{1-\frac{2}{9}\times\frac{1}{4}}\)

= \(\frac{\frac{17}{36}}{\frac{34}{36}}\)

= \(\frac{1}{2}\), true.