Question

From a window (h m high above the ground ) of a house in a street , the angle of elevation and depression of the top and the foot of another house on the opposite side of the street are α and β respectively. Show that the height of the opposite house is h(1+tan α tan β)​

Answer 1

Appropriate Question :-

From a window (h m high above the ground ) of a house in a street , the angle of elevation and depression of the top and the foot of another house on the opposite side of the street are α and β respectively. Show that the height of the opposite house is h(1 + tan α cot β)

$\large\underline{\sf{Solution-}}$

Given that,

From a window (h m high above the ground ) of a house in a street , the angle of elevation and depression of the top and the foot of another house on the opposite side of the street are α and β respectively.

Let assume that window be at the point A which is at the height of h m from the ground.

So, AB = h m

Let assume that CD be the house on the opposite side of the street.

Let assume that CD = H m

Now, from A, draw AE perpendicular to CD, intersecting CD at E.

Let assume that AE = x m

Now, In right angle triangle AEC

$\rm \: tan \beta \: = \: \dfrac{EC}{AE}$

$\rm \: tan \beta \: = \: \dfrac{h}{x}$

$\rm \: cot \beta \: = \: \dfrac{x}{h}$

$\rm\implies \:x \: = \: h \: cot \beta$

Now, In right angle triangle ADE

$\rm \: tan \alpha = \dfrac{ED}{AE}$

$\rm \: tan \alpha = \dfrac{H - h}{x}$

$\rm \: tan \alpha = \dfrac{H - h}{h \: cot \beta }$

$\rm \: H - h = h \: cot \beta \: tan \alpha$

$\rm \: H = h + h \: cot \beta \: tan \alpha$

$\rm \: H = h (1 + \: cot \beta \: tan \alpha )$

Hence, Proved

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Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$