Question

From the window (60 m high above the ground) of a house in a street, the angles of elevation and depression of the top and foot of another house on the opposite side of the street are 60° and 45° respectively. Find the height of the opposite house.
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Answer 1

S O L U T I O N :

Refer to the attached figure,

B and C the height of the opposite house is BC.

Let BC be x .

The window (60 m high above the ground) of a house in a street. [Given]

Hence AD= 60 .

Use √3 as 1.732 .

Also use tan Ф = ( side opposite to Ф ) / ( side adjacent to Ф )

Use tan 45 = 1 .

Use tan 60 = 1/√3 .

∠ACB = 60°

∠ADB = 45°

In Δ ABD ,

tan 45° = AB /BD

➮ tan 45° = 75/BD

➮ 1 = 60/BD

➮ BD = 60

From the figure we have BD + DC = BC

➮ 60 + x = BC

In Δ ABC ,

tan 60° = AB/BC

➮ 1/√3 = 60 / ( 60 - x )

➮ 60 - x = 60√3

➮ x = 60√3 + 60

➮ x = 60 ( √3 + 1 )

➮ x = 60 ( 1.732 + 1 )

➮ x = 60 × 2.732

➮ x = 163.92

Therefore, the height of the opposite house is 163.92m.

Answer 2

$\rule{200}3$

$\large{\red{\underline{\tt{Figure\::-}}}}$

$\setlength{\unitlength}{1.5cm}\begin{picture}(10,10) \thicklines \put(0,0){\line(1,0){2}}\put(0,0){\line(0,1){2}}\put(2,0){\line(0,1){4}}\put(0,2){\line(1,0){2}}\put(0,2){\line(1,-1){2}}\put(0,2){\line(1,1){2}}\put(-0.2,-0.2){ A }\put(2,-0.2){ B }\put(2.2,2){ C }\put(2.2,4){ D }\put(-0.2,2){ E }\put(-0.6,1){ 60\;m }\put(0.5,1.6){ 45^{\circ} }\put(1.38,0.15){ 45^{\circ} }\put(0.4,2.1){ 60^{\circ} }\qbezier(0.2,2.2)(0.3,2.1)(0.21,2)\qbezier(0.3,2)(0.27,1.9)(0.2,1.78)\put(2.3,3){ x }\end{picture}$

$\rule{200}3$

$\large{\red{\underline{\tt{Solution\::-}}}}$

In ∆BAE,

$\:\:$

$\dashrightarrow$ $\:$$\sf \dfrac{EA}{AB} = tan \: 45^{\circ}$

$\:\:$

$\dashrightarrow$$\:$ $\sf \dfrac{60}{y} = 1$ (Let AB = y )

$\:\:$

$\dashrightarrow$ $\:$$\sf y = 60\:m$

$\:\:$

In ∆DCE,

$\dashrightarrow$ $\:$$\sf \dfrac{DC}{CE} = tan\:60^{\circ}$

$\:\:$

$\dashrightarrow$ $\:$$\sf \dfrac{x}{60} = \sqrt{3}$

$\:\:$

$\dashrightarrow$ $\:$$\sf x = 60\sqrt{3}$

$\:\:$

Therefore, height of opptosite house = ( x + 60 )m

$\:\:$

$\dashrightarrow$ $\:$$\sf \left (60 + 60\sqrt{3} \right)m$

$\:\:$

$\dashrightarrow$$\:$ $\sf 60 \left (1 + \sqrt{3}\right)m$

$\:\:$

$\dashrightarrow$$\:$ $\sf 60 \times (1.732 + 1)$$\:\:\:\:$ $\pink{\sf (\because \sqrt{3} = 1.732)}$

$\:\:$

$\dashrightarrow$$\:$ $\sf 60 \times (2.732)$

$\:\:$

$\gray\dashrightarrow$$\:$ $\huge\underline{\boxed{\gray{\sf 169.92\:m}}}$ $\dagger$

Therefore, the height of the opposite house is 169.92m.

$\rule{200}3$