From a window ( $h$ m 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 $h$ $(1 + tan θ cot ϕ)$ m.

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Solution

$tan ϕ = \frac{h}{A C} = A C = \frac{h}{tan ϕ} = A C = h cot ϕ tan θ = \frac{B C}{A C} B C = A C tan θ = h cot ϕ tan θ$
height= $h + h cot ϕ tan θ = h (1 + cot ϕ tan θ) m$

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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 $h (1 + t a n θ c o t ϕ)$ metres [3 MARKS]

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From a window (h m 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+tanθcotϕ) m.

tanϕ=hAC=AC=htanϕ=AC=hcotϕtanθ=BCACBC=ACtanθ=hcotϕtanθheight=h+hcotϕtanθ=h(1+cotϕtanθ)m

From a window ( $h$ m 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 $h$ $(1 + tan θ cot ϕ)$ m.

$tan ϕ = \frac{h}{A C} = A C = \frac{h}{tan ϕ} = A C = h cot ϕ tan θ = \frac{B C}{A C} B C = A C tan θ = h cot ϕ tan θ$
height= $h + h cot ϕ tan θ = h (1 + cot ϕ tan θ) m$