Question

AB is a vertical pole. The end A is on the level ground. C is the middle point of AB. P is a point on the level ground. The portion BC subtends an angle β at P. If AP = n AB, then tan β is equal to

(a) \(\frac{n}{n^2-2}\)

(b) \(\frac{n}{n^2-1}\)

(c) \(\frac{n}{n^2+1}\)

(d) \(\frac{n}{2n^2+1}\)

Answer 1

 (d) \(\frac{n}{2n^2+1}\)

Let the height of the vertical pole AB = h metres. 

Then, AC = BC = \(\frac{h}{2}\), AP = n. AB = \(\frac{n}{h}\) 

Let ∠APC = α. 

Given ∠BPC = β 

∴ ∠APB = α + β 

In rt. Δ ACP,

tan α = \(\frac{AC}{AP}\) = \(\frac{\frac{h}{2}}{n\,h}\) = \(\frac{1}{2h}\)                      ....(i)

In rt. Δ BAP, tan (α + β) = \(\frac{AB}{AP}\) = \(\frac{h}{n\,h}\) = \(\frac{1}{n}\)           .....(ii)

Now tan b = tan {(α + β) – α}

= \(\frac{\text{tan}\,(\alpha+\beta)-\text{tan}\,\alpha}{1+\text{tan}\,(\alpha+\beta)\,\text{tan}\,\alpha}\)

= \(\frac{\frac{1}{n}-\frac{1}{2n}}{1+\frac{1}{n}.\frac{1}{2n}}\) = \(\frac{n}{2n^2+1}\).                  (Using (i) and (ii))