Question

The angle of depression of the top and the foot of the chimney as seen from the top of a second chimney which is 150 m high and standing on the same level as the first are θ and ϕ respectively. The distance between their tops when tan θ = \(\frac{4}{3}\)  and tan ϕ = 5/2 is equal to

(a) 50 m

(b) 100 m

(c) 100\(\sqrt2\) m

(d) \(\frac{200}{\sqrt2}\) m

Answer 1

(b) 100 m.

Let PQ and AB be the two chimneys such that the angles of depression of points A and B from point P are θ and ϕ respectively. Also, PQ = 150 m Draw AR || QB meeting PQ in R.

∴ RA = QB, RQ = AB = h (say)

Also, given,

∠PAR = θ, ∠PBQ = ϕ.

∴ In rt. Δ PRA,

tan θ = \(\frac{PR}{RA}\)

⇒ \(\frac{4}{3}\) = \(\frac{PR}{RA}\) ...(i)

In rt. D PQB,

tan ϕ = \(\frac{PQ}{QB}\)

⇒ \(\frac{5}{2}\) = \(\frac{150}{QB}\) 

⇒ QB = \(\frac{150\times2}{5}\) = 60 m ....(ii)

∵ QB = RA, putting RA = 60 in (i), we get

\(\frac{4}{3}=\frac{PR}{60}\)

⇒ PR = 80 m.

∴ In rt. Δ PRA,

distance between the top of the two chimneys, i.e.,

AP = \(\sqrt{PR^2+RA^2}\)  (Pythagoras’ Th.)

= \(\sqrt{80^2+60^2}\)

= \(\sqrt{6400+3600}\)

= \(\sqrt{10000}\)

= 100 m.