From a window (h metres high above the ground) of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are $θ$ and $ϕ$ respectively. Show that the height of the opposite house is $h (1 + t a n θ c o t ϕ)$ metres [3 MARKS]

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Solution

Let AB be the house with window at B and let CD be the another house. Then, AB = h metres.

Draw $B E ∥ A C$ , meeting CD at E. Then,

$∠ E B D = θ$ and $∠ A C B = ∠ E B C = ϕ$

Let CD = H metres. Then,

CE = AB = h metres and

ED = (H - h) m

From right $Δ A C B$ , we have

$\frac{A C}{A B} = c o t ϕ \Rightarrow \frac{A C}{h} = c o t ϕ$

$\Rightarrow A C = h c o t ϕ m e t r e s$

From right $Δ B E D$ , we have

$\frac{D E}{B E} = t a n θ \Rightarrow \frac{(H - h)}{h c o t ϕ} = t a n θ$ $[∵ B E = A C = h c o t ϕ m]$

$\Rightarrow (H - h) = h t a n θ c o t ϕ$

$\Rightarrow H = h (1 + t a n θ c o t ϕ)$

Hence, the height of the opposite house is $h (1 + t a n θ c o t ϕ)$ metres.

Alternative Method,

[ $\frac{1}{2}$ MARK]

In $Δ A B C$

$t a n θ = \frac{y}{x}$

$x = \frac{y}{t a n θ} \dots \dots (1)$ [1 MARK]

In $Δ A C D$

$t a n ϕ = \frac{h}{x}$

$t a n ϕ = \frac{h}{\frac{y}{t a n θ}}$ (from (1)) [1 MARK]

$t a n ϕ = \frac{h t a n θ}{y}$

$y = \frac{h t a n θ}{t a n ϕ}$ (2)

Height of opposite house

= y + h

$= \frac{h t a n θ}{t a n ϕ} + h$

$= \frac{h (t a n θ + t a n ϕ)}{t a n ϕ}$

$= h (t a n θ c o t ϕ + 1) m e t r e s$ [1 $\frac{1}{2}$ MARK]

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From a window (h metres high above the ground) of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are θ and ϕ respectively. Show that the height of the opposite house is h(1+tan θ cot ϕ) metres [3 MARKS]