the curved surface area of a cylinder is 4400 cm2 and the circumference of its base 110 cm find the volume of the cylinder
Answers
Step-by-step explanation:
\begin{gathered}{\bf{Given\::}} \\ \end{gathered}
Given:
The angle of elevation of the top of a building from foot of the tower is 30°.
The angle of elevation of the top of the tower from the root of the building is 45°.
Height of the tower is 30 m.
\begin{gathered} \\ {\bf{To\: Find\::}} \\ \end{gathered}
ToFind:
The height of the building.
\begin{gathered} \\ {\bf{Solution\::}} \\ \end{gathered}
Solution:
Let,
As shown in figure,
Height of the building = PQ.
Height of the tower = RS = 30 m.
Angle of elevation of the top of a building from foot of the tower = \rm{\angle{PRQ}}∠PRQ = 30°
Angle of elevation of the top of the tower from the root of the building = \rm{\angle{SQR}}∠SQR = 45°
Since,
Building & tower are perpendicular to the ground.
\longrightarrow\:\bf{\angle{PQR}\:=\:\angle{SRQ}\:=\:90^{\circ}\:}⟶∠PQR=∠SRQ=90
∘
Now,
In right angle ∆SQR,
➠ \tt{\tan{45^{\circ}}\:=\:\dfrac{SR}{QR}\:}tan45
∘
=
QR
SR
➠ \tt{1\:=\:\dfrac{30}{QR}\:}1=
QR
30
➠ \bf{QR\:=\:30\:cm}QR=30cm
Again,
In right angle ∆PQR,
➠ \begin{gathered}\tt{\tan{30^{\circ}}\:=\:\dfrac{PQ}{30}\:} \\ \end{gathered}
tan30
∘
=
30
PQ
➠ \begin{gathered}\tt{\dfrac{1}{\sqrt{3}}\:=\:\dfrac{PQ}{30}\:} \\ \end{gathered}
3
1
=
30
PQ
➠ \tt{PQ\:=\:\dfrac{1}{\sqrt{3}}\times{30}\:}PQ=
3
1
×30
➠ \bf\pink{PQ\:=\:10\sqrt{3}\:cm\:=\:17.32\:cm}PQ=10
3
\begin{gathered}{\bf{Given\::}} \\ \end{gathered}
Given:
The angle of elevation of the top of a building from foot of the tower is 30°.
The angle of elevation of the top of the tower from the root of the building is 45°.
Height of the tower is 30 m.
\begin{gathered} \\ {\bf{To\: Find\::}} \\ \end{gathered}
ToFind:
The height of the building.
\begin{gathered} \\ {\bf{Solution\::}} \\ \end{gathered}
Solution:
Let,
As shown in figure,
Height of the building = PQ.
Height of the tower = RS = 30 m.
Angle of elevation of the top of a building from foot of the tower = \rm{\angle{PRQ}}∠PRQ = 30°
Angle of elevation of the top of the tower from the root of the building = \rm{\angle{SQR}}∠SQR = 45°
Since,
Building & tower are perpendicular to the ground.
\longrightarrow\:\bf{\angle{PQR}\:=\:\angle{SRQ}\:=\:90^{\circ}\:}⟶∠PQR=∠SRQ=90
∘
Now,
In right angle ∆SQR,
➠ \tt{\tan{45^{\circ}}\:=\:\dfrac{SR}{QR}\:}tan45
∘
=
QR
SR
➠ \tt{1\:=\:\dfrac{30}{QR}\:}1=
QR
30
➠ \bf{QR\:=\:30\:cm}QR=30cm
Again,
In right angle ∆PQR,
➠ \begin{gathered}\tt{\tan{30^{\circ}}\:=\:\dfrac{PQ}{30}\:} \\ \end{gathered}
tan30
∘
=
30
PQ
➠ \begin{gathered}\tt{\dfrac{1}{\sqrt{3}}\:=\:\dfrac{PQ}{30}\:} \\ \end{gathered}
3
1
=
30
PQ
➠ \tt{PQ\:=\:\dfrac{1}{\sqrt{3}}\times{30}\:}PQ=
3
1
×30
➠ \bf\pink{PQ\:=\:10\sqrt{3}\:cm\:=\:17.32\:cm}PQ=10
3
cm=17.32cm
∴ Height of the building is 17.32 cm.
cm=17.32cm
∴ Height of the building is 17.32 cm.