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 \[\sqrt{3}=1.732\]).
Last updated date: 18th Mar 2023
•
Total views: 306.6k
•
Views today: 3.85k
Answer
306.6k+ 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.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
