# From a window 15 meters height above the ground 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 \[{{30}^{{}^\circ }}\] and \[{{45}^{{}^\circ }}\]respectively, show that the height of the opposite house is 23.66 meters. (Take \[\sqrt{3}=1.732\]).

Answer

Verified

364.5k+ views

Hint: First, construct an appropriate diagram with the given information. Assume the height of the building that you have to find as ‘4’. Finally, determine the expression for \[tan\theta \] from the two triangles and obtain the value of ‘h’ accordingly.

It was given that the angles of elevation and depression of the top and foot of a different house on the opposite side of the street are \[30{}^\circ \]and \[45{}^\circ \]respectively from a height of 15m above the ground.

First, let us understand the basic difference between the angle of elevation and angle of depression.

For suppose, if you were looking at something above the horizon, then the angle made by the horizontal and your line of right is the angle of elevation. Similarly, if you were looking at something that is below the horizon, then we are looking at something that is below the horizon, then we are looking at something that is below the horizon, then the angle made by the horizontal and your line of sight is the angle of depression.

Now let us construct the diagram with information we have:

We have assumed the height of another building is ’h’, now from \[\Delta PQC\], we have:

\[tan\theta \]= opposite side / hypotenuse

\[tan{{45}^{{}^\circ }}=\dfrac{QC}{PQ}\]

From figure we have QC = PA = 15 and substituting the value of ‘tan’, we get

\[1=\dfrac{15}{PQ}\]

\[PQ=15m\]

Now, in \[\Delta PQC\], we have:

QD = CD – QC

Substituting the values from the figure, we get

\[QD=\left( h-15 \right)m.......(i)\]

Now as we know, \[tan\theta \]= opposite side / hypotenuse

\[\Rightarrow tan{{30}^{{}^\circ }}=\dfrac{QD}{PQ}\]

Substituting the value of ‘tan’ and value from equation (i), we get

\[\dfrac{1}{\sqrt{3}}=\dfrac{h-15}{15}\]

\[h-15=\dfrac{15}{\sqrt{3}}\]

\[\begin{align}

& h=\dfrac{15}{\sqrt{3}}+15 \\

& \Rightarrow h=\dfrac{(5\times 3)\sqrt{3}}{\sqrt{3}\times \sqrt{3}}+15 \\

\end{align}\]

Solving further we get:

\[h=5\sqrt{3}+15\]

Taking ‘5’ common, we have:

\[h=5\left( \sqrt{3}+5 \right)\]

Substituting \[\sqrt{3}=1.732\] in the above equation, we get:

\[h=5\left( 3+1.732 \right)\]

\[h=5\left( 4.732 \right)\]

\[h=23.66\] Meters

Hence, we have proved that the height of the opposite house is \[h=23.66\] meters.

Note: Constructing a proper diagram is mandatory for these questions. You may also go ahead by considering the expressions of \[\sin \theta \] and \[\cos \theta \] from the triangle, but it would require extra effort to solve.

It was given that the angles of elevation and depression of the top and foot of a different house on the opposite side of the street are \[30{}^\circ \]and \[45{}^\circ \]respectively from a height of 15m above the ground.

First, let us understand the basic difference between the angle of elevation and angle of depression.

For suppose, if you were looking at something above the horizon, then the angle made by the horizontal and your line of right is the angle of elevation. Similarly, if you were looking at something that is below the horizon, then we are looking at something that is below the horizon, then we are looking at something that is below the horizon, then the angle made by the horizontal and your line of sight is the angle of depression.

Now let us construct the diagram with information we have:

We have assumed the height of another building is ’h’, now from \[\Delta PQC\], we have:

\[tan\theta \]= opposite side / hypotenuse

\[tan{{45}^{{}^\circ }}=\dfrac{QC}{PQ}\]

From figure we have QC = PA = 15 and substituting the value of ‘tan’, we get

\[1=\dfrac{15}{PQ}\]

\[PQ=15m\]

Now, in \[\Delta PQC\], we have:

QD = CD – QC

Substituting the values from the figure, we get

\[QD=\left( h-15 \right)m.......(i)\]

Now as we know, \[tan\theta \]= opposite side / hypotenuse

\[\Rightarrow tan{{30}^{{}^\circ }}=\dfrac{QD}{PQ}\]

Substituting the value of ‘tan’ and value from equation (i), we get

\[\dfrac{1}{\sqrt{3}}=\dfrac{h-15}{15}\]

\[h-15=\dfrac{15}{\sqrt{3}}\]

\[\begin{align}

& h=\dfrac{15}{\sqrt{3}}+15 \\

& \Rightarrow h=\dfrac{(5\times 3)\sqrt{3}}{\sqrt{3}\times \sqrt{3}}+15 \\

\end{align}\]

Solving further we get:

\[h=5\sqrt{3}+15\]

Taking ‘5’ common, we have:

\[h=5\left( \sqrt{3}+5 \right)\]

Substituting \[\sqrt{3}=1.732\] in the above equation, we get:

\[h=5\left( 3+1.732 \right)\]

\[h=5\left( 4.732 \right)\]

\[h=23.66\] Meters

Hence, we have proved that the height of the opposite house is \[h=23.66\] meters.

Note: Constructing a proper diagram is mandatory for these questions. You may also go ahead by considering the expressions of \[\sin \theta \] and \[\cos \theta \] from the triangle, but it would require extra effort to solve.

Last updated date: 29th Sep 2023

•

Total views: 364.5k

•

Views today: 5.64k

Recently Updated Pages

What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is meant by shramdaan AVoluntary contribution class 11 social science CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

An alternating current can be produced by A a transformer class 12 physics CBSE

What is the value of 01+23+45+67++1617+1819+20 class 11 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers