Angle of elevation:-
Here, " the line of sight", "horizontal line" and the tree form a right angle triangle.
The line of sight is above the horizontal line and angle between the line of sight and the horizontal line is called angle of elevation.
Angle of depression:
Suppose you are standing on the top of your school building and you want to find the distance of borewell from the building on which you are standing. For that, you have observe the base of the borewell.
Then, the line of sight from your eye to the base of borewell is below the horizontal line from your eye.
Here, "the angle between the line of sight and horizontal line is called angle of depression"
Example-1. The top of a clock tower is observed at angle of elevation of π and the foot of the tower is at the distance of d meters from the observer. Draw the diagram for this data.
sol) The diagrams are as shown below.
******************************************************************************
Example-2. Rinky observes a flower on the ground from the balcony of the first floor of a building at an angle of depression π·. The height of the first floor of the building is x meters. Draw the diagram for this data.
sol)
Here ∠DAC = ∠ACB = π·
*****************************************
Example-3. A large balloon has been tied with a rope and it is floating in the air. A person has observed the balloon from the top of a building at angle of elevation π½1 and foot of the rope at an angle of depression of π½2.
The height of the building is h feet. Draw the diagram for this data.
sol) We can see that
∠BDA = ∠DAE
***************************************
Example-4. A boy observed the top of an electric pole at an angle of elevation of 60 deg when the observation point is 8 meters away from the foot of the pole. Find the height of the pole.
sol) From the figure, in triangle OAB
OB = 8 meters
AOB = 60 deg
Let, height of the pole=AB=h meters
(we know the adjacent side and we need to find the opposite side of AOB in the triangle OAB.
Hence we need to consider the trigonometric ratio "tan" to solve the problem).
tan 60 = AB
OB
3 = h
8
h = 8*3 m.
****************************************
Example-5. Rajender observes a person standing on the ground from a helicopter at an angle of depression 45 deg. If the helicopter flies at a height of 50 meters from the ground, what is the distance of the person from Rajender?
sol) From the figure, in triangle OAB
OA = 50 meters
POB = OAB =45 deg
OB = distance of the person from rajender=x
(We know the opposite side of OBA and we need to find hypotenuse Ob in the triangle OAB. Hence, we need to consider the ration "sin".)
sin 45 = OA
OB
1 = 50
2 x
x = 50* 2 meters
( The distance from the person to Rajender is 50 2 m).
*****************************************
Exercise - 12.1
(1) A tower stands vertically on the ground. From a point which is 15 meter away from the foot of the tower, the angle of elevation of the top of the tower is 45 deg. What is the height of the tower?
sol) Let, the height of the tower be "H" meter
So, AB = H meter
Since tower is vertical to ground ,
So, /_ABC = 90 deg
From the figure, in right angled triangle ABC
Distance between a point and a tower (CB) = 15 mts
Angle of elevation = 45 deg
/_ACB = 45 deg
RTP :- AB = height of the tower
We know the adjacent side of /_ABC and we need to find opposite side AB in the traingle ABC. Hence we need to consider the ratio of "tan"
Now,
tan 45 deg = opposite--
Adjacent
" = AB
CB
************
Tan 45 deg = 1
************
1 = H
15
H = 15 meter. is the height of the tower.
**********************************
(2) A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making 30 deg angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is 6 m. Find the height of the tree before falling down.
sol) Let the height of a tree before broken is "X" meters
Given :- Tree touches the ground by making 30 deg angle
Given :- Distance between foot and top of a tree(AB) = 6 m
RTP :- Height of the tree before falling
From the figure, in right angled triangle OAB
AB = 6 m
/_AOB = 30 deg
We know the opposite side of /_OAB and we need to find adjacent side OB in the triangle OAB. Hence we need to consider the ratio of "tan"
tan 30 deg = AB
OB
1 = 6
_/3 OB
OB = 6 _/3 m
H = 6 _/3 is he height of tree before falling.
*************************************
(3) A contractor wants to set up a slide for the children to play in the park. He wants to set it up at the height of 2m and by making an angle of 30 deg with the ground. What should be the length of the slide?
sol) Given :- Height(h) = 2 m and0 = 30
RTP :- OA = ?
From the figure, in triangle OAB
OB = 2 m
/_ OAB = 30 deg
sin 30 = OB
OA
1 = 2
2 OA
OA = 4 m
X = 4 m is the length of the slide
*****************************
(4) length of the shadow of a 15 meter high pole us 5 _/3 meters at &'o clock in the morning. Then what is the angle of elevation of the Sun rays with the ground at the time?
sol) Let Angle of elevation of the Sun rays with the ground /_ACB be "x"
Height of the pole = AB = 15 m
Length of the shadow = BC = 5_/3 m
tan x =15
5_/3
= 3
_/3
= _/3 * _/3
-------------- (_/3 *_/3 = 3 )
_/3
x = _/3
Angle of elevation = 60 deg ( _/3 = tan 60)
**************************************
(5) You want to erect a pole of height 10 m with the support of three ropes. Each rope has to make an angle 30 deg with the pole. What should be the length of the rope?
sol) From the figure
In /_\ PQR,
Height of the pole(QR) = 10
Length of rope (PR) = ?
/_PRQ = 30 deg
We Know :- The adjacent side of /_PRQ and we need to find hypotenuse "PR" in the triangle PQR. Hence, we need to consider the ratio "cos"
cos 30 = QR
PR
_/3 = QR
2 PR
_/3 = 10
2 PR
PR = 20 (_/3 = 1.73)
_/3
PR = 11.56 m is the length of the rope
************************************
(6) Suppose you are shooting an arrow from the top of a building at an height of 6m to a target on the ground at an angle of depression of 60 deg. What is the distance between you and the object?
sol) From the figure, In triangle ABC
Height of the building ( OA) = 6 m
/_BAC = 60 deg
Distance between you and object (AC ) = x
We know the opposite side of /_BAC and we need to find hypotenuse "AC" in the triangle ABC. Hence, we need to consider the ratio "sin"
sin 60 = Opposite
Hypotenuse
sin 60 = BC
AC
_/3 = 6
2 AC
AC = 12
_/3
AC = 4 * 3
_/3
AC = 4 *_/3 * _/3
------------------
_/3
AC = 4 _/3 m is the distance between you and building
***********************************************
(7) An electrician wants to repair an electric connection on a pole of height 9m. He needs to reach 1.8 m below the top of the pole to do repair work. What should be the length of the ladder which he should use. When he climbs it an angle of 60 deg with the ground? What will be the distance between foot of the ladder and foot of the pole?
sol) From the figure,,
height of the pole (PB) = 9 cm
AB = 9 - 1.8 = 7.2 cm
/_BCA = 60 deg
Distance between the foot of ladder and foot of pole (BC) = ?
tan 60 = AB
BC
_/3 = 7.2
BC
BC = 7.2
1.73
BC = 4.161 m
**********************************
(8) A boat has to cross a river. It crosses the river by making an angle of 60 deg with the bank of the river due to the stream of the river and travels a distance of 600 m to reach the another side of the river. What is the width of the river?
sol) Let boat starts from "C" and reaches at "A".
/_ACB = 60 deg
AC = 600 m
cos 60 = BC
AC
1 = BC
2 600
BC = 300 m is the width of the river.
****************************************
(9) An observer of height 1.8 m is 13.2 m away from a palm tree. The angle of elevation of the top of the tree from his eyes is 45 deg. What is the height of the palm tree?
sol) From figure OAB
Distance between observer and a tree (AB) = 13.2 m
/_OBA = 45 deg
(WKT the adjacent side of /_OBA and we need to find opposite side OA in the triangle OAB.Hence , we need to consider the ratio "tan" )
tan 45 = Opposite side
Adjacent Side
tan 45 = OA
AB
1 = OA
13.2
OA = 13.2 m
The height of palm tree is OA + AD = 13.2 + 1.8 = 15 m
*************************************************
(10) In the adjacent figure, AC = 6cm, AB = 5 cm and /_BAC = 30 deg. Find the area of the triangle.
sol) In a /_\ABC, draw BD as perpendicular to AC
⧍ BAD is a right angle triangle.
sin 30 = opposite side
Hypotenuse
sin 30 = BD
AB
sin 30 = x
5
sin 30 * 5 = x
1 * 5 = x
2
5 = x
2
2.5 = x
opposite side (x) or Height of /_\ BAD= 2.5 cm
Area of /_\ ABC = 1/2 * 2.5 * 6
=> 7.5 cm
********************************************************************************
*
(*) Solution for Two triangles
Suppose you are observing the top of the palm tree at an angle of elevation 45 deg. The angle of elevation changes to 30 deg when you move 11m away from the tree.
Let us see how we can find height of the tree.
From figure, we have
AB = 11 m
∠DAC = 30 deg
∠DBC = 45 deg
Let the height of the palm tree CD = h meters
and length of BC = x
AC = 11 + x
Case(!): Traingle BDC
Tan 45 = DC
BC
1 = h
x
x = h ----------(1)
Case (!!): triangle ADC
Tan 30 = DC
AC
1 = h
√3 11 + x
h = 11 + x
√3
h = 11 + x
√3 √3
h = 11 + h
√3 √3
h - h = 11
√3 √3
h (√3 - 1 ) = 11
√3 √3
h = 11 meters
(√3 - 1)
NOTE: Total height of the palm tree is
CD + CE
where CE = AF, which is the height of the girl.
*****************************************
Example-6. Two men on either side of a temple of 30 meter height observe its top at the angles of elevation 30 deg and 60 deg respectively. Find the distance between the two men.
sol) Height of the temple BD = 30 deg
Angle of elevation of one person ∠BAD = 30
Angle of elevation of another person ∠BCD = 60
Let the distance between the first person and the temple ,
AD = x and distance between the second person and the temple,
CD = d
From △ BAD
Tan 30 = BD
AB
1 = 30
√3 x
x = 30 √3-----(1)
(*) From △ BCD
Tan 60 = BD
d
√3 = 30
d
d = 30 ----(2)
√3
from (1) and (2) distance between the persons = BC + BA = x + d
= 30 √3 + 30
√3
= 30 *3 + 30
√3
= 90 +30
√3
= 120
√3
= 120 √ 3
√3* √ 3
= 120√ 3
3
= 40√3 meter.
************************************
*****************************************
Example-7. A straight highway leads to the foot of a tower. ramaiah standing at the top of the tower observes a car at an angle of depression 30 deg.
The car is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60 deg.
Find the time taken by the car to reach the foot of the tower from this point.
sol) Let the distance travelled by the car in 6 seconds = AB = x meters
height of the tower CD = h meters
The remaining distance to be travelled by the car BC = d meters
and AC = AB + BC= (x +d) meters
∠ PDA = ∠ DAP = 30
∠PDB = ∠ DBP = 60
Case(!) : Triangle BCD
Tan 60 = CD
BC
√3 = h
d
h = √3d ---------(1)
case(!!) Triangle ACD
Tan 30 = CD
AC
1 = h
√3 (x+d)
h = (x +d) ---------(2)
√3
From (1) and (2), we have
x +d = √3d
√3
x + d = 3d
x = 2d
x = d
2
Time taken to travel 'x' meters = 6 seconds.
Time taken to travel the distance of 'd' meters
i.e,
x meters = 6 seconds.
2 3
= 3 seconds.
****************************************
Exercise- 12.2
(1) A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower , the angle of elevation of the top of tower is 60 deg. from another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30 deg. Find the height of the tower and the width of the road.
sol) Let Height (h) of tower = AB
Given :- From a point on the other side directly opposite the tower, the angle of elevation of the tower is 60 deg
Hence, /_ACB = 60 deg
Also,
Angle of elevation from point "D" to top of the tower = 30 deg
So, /_ADB = 30 deg
Also, DC = 10 m
RTP :- we need to find height of tower "AB" and width of road "BC"
So, /_ABD = 90 deg
(1) In right angle triangle ACB:-
tan C = Side opposite to angle C
Side adjacent to angle C
tan 60 = AB
BC
_/3 = AB
BC
_/3 * BC = AB
AB = _/3BC -------(1)
(2) In a right angle triangle ADB :-
tan D = Side opposite to angle D
Side adjacent to angle D
tan 30 = AB
BD
1 = AB
_/3 BD
BD = AB --------------(2)
_/3
comparing (1) and (2)
_/3BC = BD
_/3
_/3 * _/3BC = BD
3BC = BC + CD
3BC = BC + 10
3BC - BC = 10
2BC = 10
BC = 5
or
x = 5m
Hence, the width of the road is 10 m
From (1)
AB = _/3BC
AB = _/3 * 5
AB = 5_/3 m is the height of the tower
**************************************
(2) A 1.5 m tall boy is looking at the top of a temple which is 30 meter in height from a point at certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30 deg to 60 deg as he walks towards the temple. Find the distance he walked towards the temple.
sol) From figure,
/_DAC = 30 deg and
/_DBC = 60 deg
Since DE = 30 m,
height AF // CE
AF = CE = 1.5 m
CD = DE -- CE
= 30 - 1.5
= 28.5 m
(!) From /_\ DAC
tan 30 = CD
AC
=> 1 = 28.5
_/3 AC
AC = 28.5 * _/3 m {√3 = 1.73 or 2}
AC = 49.36 -----------(1)
(2) From /_\ DBC
tan 60 = DC
BC
=> _/3 = 28.5
BC
=> BC = 28.5
_/3
=> BC = 16.45 ----------------(2)
Distance he walked towards temple = AB
AB = AC -- BC
AB = (1) --- (2)
AB = 49.36 -- 16.43
AB = 32.93 m is the distance he walked towards the temple.
***************************************************
(3) A statue stands on the top of a 2m tall pedestal. From a point on the ground , the angle of elevation of the top of the statue is 60 deg and from the same point, the angle of elevation of the top of the pedestal is 45 deg. Find the height of the statue.
sol) From the figure,
/_DBC = 45 deg and
/_ABC = 60 deg
The height of pedestal (DC) = x m
The height of statue(AD) = 2 m
from /_DBC = 45 deg
tan 45 = DC
BC
1 = DC
BC
BC = DC --------(1)
(2) /_\ ABC = 60 deg
tan 60 = AC
BC
_/3 = AD + DC
BC
_/3 = 2 + DC
DC -----------> ( BC= DC)
DC_/3 = DC + 2
DC√3 - DC = 2
DC (√3 - 1) = 2
DC = 2
(√3 - 1)
DC = 2
(1.73 -1) (√3 = 1.73)
DC = 2
0.73
DC = 200
73
DC = 2.73 m is the height of the pedestal
*********************************
(4) From the top of a 7m building, the angle of elevation of the top of a cell tower is 60 deg and the angle of depressions to its foot is 45 deg. find the height of the tower.
sol) From the figure
Given :- Angle of elevation ( /_EAD) = 60 deg
Angle of depressions ( /_ACD ) = 45 deg
Since AB & CD are parallel
DC = AB = 7m
Also, AD & BC are parallel
So, AD = BC
taking AC as traversal
∠ ACB = ∠ DAC
∠ACB = 45 deg
R.T.P :- Height(CE)=?
(1) From /_ABC
tan 45 = AB
BC
1 = 7
BC
BC = 7 m
Since BC = AD
Also, AD = 7m
(2) From /_EAD
tan 60 = ED
AD
_/3 = ED
7
ED = (7 * _/3) = (7 * 1.73) = 12. 11 m
Height of the tower is :-
=> ED + DC
=> 12.11 + 7
=> 19. 12 m
***************************************************
(5) A wire of length 18m had been tied with electric pole at an angle of elevation 30 deg with the ground. Because it was covering a long distance, it was cut and tied at an angle of elevation 60 with the ground. . How much length of the wire was cut?
sol) From the figure
(1) In a /_\ ABC,
AB = 18
/_ABC = 30 deg
sin 30 = AC
AB
1 = AC
2 AB
1 = AC
2 18
AC = 9 m.
(2) In a /_\ADC
AC = 9 m
/_ADC = 60 deg
sin 60 = AC
AD
√3 = 9
2 AD
AD = 18
_/3
Length of the wire cut = AB -- AD
=> 18 -- 18
_/3
=> 18 -- 18
1.73
=> 18 -- 10.40
=> 7.61 m is the length of the wire had been cut.
******************************************************
(6) The angle of elevation of the top of a building from the foot of the tower is 30 deg and the angle of elevation of the top of the tower from the foot of the building is 60 deg. If the tower is 30 m high, find the height of the building.
sol) From the figure
/_ACB = 60 deg and
/_DBC = 30 deg
Height of tower(AB) = 30 m
Height of building (DC) = x
(1) From /_ACB
tan 60 = AB
BC
_/3 = 30
BC
BC = 10_/3
BC = 10 * 1.73
BC = 17.3 m
(2) From /_DBC
tan 30 = DC
BC
1 = DC
_/3 10_/3
DC = 10m is the height of building
*********************************** *****************
(7) Two poles of equal heights are standing opposite to each other on either side of the road which is 120 feet wide. From a point between them on the road , the angles of elevation of the top of the pole are 60 deg and 30 deg respectively. Find the height of the poles and the distances of the point from the poles.
sol) From the figure
(1) In DPC
tan 30 = CD
CP
1 = CD
_/3 CP
or
CD = CP ------(1)
_/3
(2) In /_\ APB
tan 60 = AB
PB
_/3 = AB
PB
_/3 BP = AB
CD = √3 BP---(2) ( Pole AB = Pole CD)
equate (1) and (2)
we get
CP = √3 BP
√3
CP = 3 BP (√3 * √3 = 3)
Now,
BC = BP + CP
120 = BP + 3BP (CP = 3BP)
120 = 4BP
30 = BP
Now,
CP = BC - BP
CP = 120 - 30
CP = 90 m
From (2)
CD = √3 BP
CD = √3 * 30
CD = 1.73 * 30
CD = 51.9 is the length of the pole
Distance of Point(p) from pole CD = CP = 90m
Distance of Point(p) from pole AB = BP = 30 m
***************************************************
(8) The angles of elevation of the top of a tower from two points at a distance of 4m and 9m . find the height of the tower from the base of the tower and in the same straight line with it are complementary.
sol) From the figure
Given :- Angles are complementary.i.e,
π +π· = 90
π· = ( 90 - π)
(1) In /_\ ABD
tan π = AB
DB
tan π = AB -----------(1)
9
(2) In /_\ ABC
Tan π· = ( AB )
BC
**************
π +π· = 90 deg
π· = (90 -π)
**************
tan (90- π) = AB
BC
cotπ = AB
4
*************
We know
cot0 = 1
tan0
*************
1 = AB
tan π 4
tan π = 4 ----------(2)
AB
equating (1) & (2)
AB = 4
9 AB
AB^2 = 9*4
AB^2 = 36
AB^2 = √36 (6*6 or 6^2 =36)
AB = 6m is the height of the tower.
****************************************************
(9) The angle of elevation of a jet plane from a point A on the ground is 60 deg. After a flight of 15 seconds , the angle of elevation changes to 30 deg. If the jet plane is flying at a constant height of 1500_/3 meter, find the speed of the jet plane ( _/3 = 1.732}
sol) From the figure
/_BAE = 60 and /_CAD = 30
(1) In /_CAD = 30
tan 30 = CD
AD
1 = 1500 _/3
_/3 AD
√3 *√3 = 1500
AD
3 = 1500
AD
AD = 3 * 1500
AD = 4500
(2) In /_BAE = 60
tan 60 = BE
AE
_/3 = 1500_/3
AE
AE = 1500
From figure
BC // ED or BC = ED
ED = AD -- AE
=> 4500 -- 1500
=> 3000m is the distance traveled from point(E) to point(D) in 15 seconds
We have
Speed = distance
Time
=> 3000
15
=> 200 m/s is the speed of the plane
***************************************************
(*) Optional Exercise
1) A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60 deg. After some time, the angle of elevation reduces to 30 deg. Find the distance traveled by the balloon during the interval.
sol) Given :- 1.2 m tall girl spots a balloon at a height of 88.2 m
The height(DE) of the balloon from the eyes of a girl will be
DE = DG - EG
DE = 88.2 - 1.2 = 87 m
Case(!) : In ◿DCE
Tan 30 = DE
CE
1 = 87
√3 CE
CE = 87 √3
Case (!!) : In ◿ABC
Tan 60 = AB
BC
√3 = 87
BC
BC = 87
√3
Multiply and divide by "√3"
BC = 87 √3
√3* √3
BC = 87 √3
3
BC = 29 √3
From the figure
Distance traveled (BE) = CE - BC
=> 87 √3 - 29 √3
=> √3 ( 87 -29)
=> 58√3 m.
*******************************************************
2) The angle of elevation of the top of a tower from the foot of the building is 30 deg and the angle of elevation of the top of the building from the foot of the tower is 60 deg. What is the ratio of heights of tower and building.
sol)
Height of tower(h1) = AB
Height of building(h2) = CD
Distance between tower and building(BC) = x meters
Case(!) : △ ABC,
Tan 30 = AB
BC
1 = h1
√3 x
x = h1 √3 ----------------(1)
Case(!!) : in △BCD
Tan 60 = CD
BC
√3 = h2
x
x = h2 ---------(2)
√3
comparing equation(1) and (2) we get
h1√3 = h2
√3
h1 = 1
h2 √3 *√3
h1 = 1
h2 3
.^., Ratio of height of tower and building
h1 : h2 = 1 : 3
***********************************************
3) The angles of elevation of the top of a lighthouse from 3 boats A,B and C in a straight line of same side of the light house are a, 2a, 3a respectively. If the distance between the boats A and B is "x" meters. Find the height of light house?
sol)The angles of elevation of the top of the lighthouse from the boats
∠PCQ = 3a
∠PBQ = 2a
∠PAQ = a
The distance between the boats A and B = x meters
The height of the lighthouse PQ = ?
Case(!) : ◿ PCQ
tan 3a = PQ
QC
tan 3a = h
QC
QC = h
tan 3a --------(1)
Case(!!) : In◿ PBQ
Tan 2a = PQ
QB
Tan 2a = h
QB
QB = h
Tan 2a ------(2)
Case (!!!) : In ◿ PAQ
Tan a = PQ
QA
Tan a = h
QA
QA = h
Tan a -------(3)
********************
But, from figure
QA = QC + BC + AB
*********************
QC + BC + AB = h
tan a
Given : AB = x mts
Now,
QC + BC + x = h
tan a
******************
from figure
BC = QB - QC
***************
QC + QB - QC + x = h
tan a
QB + x = h
tan a
Substitute(2) in above
h + x = h
Tan 2a Tan a
h + x Tan 2a = h
Tan 2a Tan a
cross multiply , we get
Tan a *( h + x Tan 2a) = h Tan 2a
h Tan a + x Tan a Tan 2a = h Tan 2a
h Tan 2a - h Tan a = x Tan a * Tan 2a
h ( Tan 2a - tan a) = x Tan a * tan 2a
h = x * Tan a * Tan 2a
Tan 2a - Tan a is the height of lighthouse
******************************************************
4) Inner part of a cupboard is in the cuboidal shape with its length, breadth and height in the ratio 1: √2 : 1. What is the angle made by the longest stick which can be inserted cupboard with its base inside.
sol) Given :- Inner part of a cupboard is in the cuboidal shape
Length(BC) = 1
Breadth(AB) = √2
Height(AE) = 1
We know that longest stick will be along the body diagonal of the cupboard.
Longest stick in the cupboard= EC
In △ABC
Using Pythagoras Theorem
AC^2 = AB^2 + BC^2
AC^2 = (√2)^2 + (1)^2
AC^2 = 2 +1
AC = √3 ----------(1)
In △ACE
Angle made by the longest stick with the base of the cupboard is given by,
Tan π½ = height of the cupboard
Diagonal of the base of cupboard
Tanπ½ = AE
AC
=> 1
√3 <----(from(1)
Tan π½ = 30 deg
.^., the angle made by the stick with the base is 30 degrees.
******************************************************
5) An iron spherical ball of volume 232848 cm^3 has been melted and converted into a cone with the vertical angle of 120 deg. What are its height and base?
sol) Given : Spherical ball Volume = 232848 cm^3
vertical angle = 120 deg
Let the radius of the base = r
Let the height of the cone = h
Since vertical angle is 120 deg
Height divides this angle into half 60 deg
r = Tan 60
h
r = h√ 3 ----(1)
****************************
We have
Volume of Cone = 1 π *r^2 *h
3
*****************************
Since the iron spherical ball converted into a cone BUT VOLUME is same .
1 * π * r^2 * h = 232848
3
1 * π * (h √3)^2 * h = 232848
3
3 *π
*h^3 = 232848
3
πΏ*h^3 = 232848
h^3 = 232848
πΏ
h^3 = 232848
(22/7 )
h^3 = 232848 * 7
22
h^3 = 74088
h^3 = 42*42*42 = 42^3
h = 42
So, the height of the cone is 42 cm.
put h = 42 in (1)
we get
r = 42* √3
r = 72.74
Finding Base :
Base = πΉ * r^2
=> πΏ * (72.74)^2
=> 3.14 * (72.74)^2
=> 16614.077 cm^2
**************************
100 cm = 1 m
10000 square cm = 1 square m
**************************
=> 1.6614 m^2
*******************************************************
Here, " the line of sight", "horizontal line" and the tree form a right angle triangle.
The line of sight is above the horizontal line and angle between the line of sight and the horizontal line is called angle of elevation.
Angle of depression:
Suppose you are standing on the top of your school building and you want to find the distance of borewell from the building on which you are standing. For that, you have observe the base of the borewell.
Then, the line of sight from your eye to the base of borewell is below the horizontal line from your eye.
Here, "the angle between the line of sight and horizontal line is called angle of depression"
Example-1. The top of a clock tower is observed at angle of elevation of π and the foot of the tower is at the distance of d meters from the observer. Draw the diagram for this data.
sol) The diagrams are as shown below.
******************************************************************************
Example-2. Rinky observes a flower on the ground from the balcony of the first floor of a building at an angle of depression π·. The height of the first floor of the building is x meters. Draw the diagram for this data.
sol)
Here ∠DAC = ∠ACB = π·
*****************************************
Example-3. A large balloon has been tied with a rope and it is floating in the air. A person has observed the balloon from the top of a building at angle of elevation π½1 and foot of the rope at an angle of depression of π½2.
The height of the building is h feet. Draw the diagram for this data.
sol) We can see that
∠BDA = ∠DAE
***************************************
Example-4. A boy observed the top of an electric pole at an angle of elevation of 60 deg when the observation point is 8 meters away from the foot of the pole. Find the height of the pole.
sol) From the figure, in triangle OAB
OB = 8 meters
AOB = 60 deg
Let, height of the pole=AB=h meters
(we know the adjacent side and we need to find the opposite side of AOB in the triangle OAB.
Hence we need to consider the trigonometric ratio "tan" to solve the problem).
tan 60 = AB
OB
3 = h
8
h = 8*3 m.
****************************************
Example-5. Rajender observes a person standing on the ground from a helicopter at an angle of depression 45 deg. If the helicopter flies at a height of 50 meters from the ground, what is the distance of the person from Rajender?
sol) From the figure, in triangle OAB
OA = 50 meters
POB = OAB =45 deg
OB = distance of the person from rajender=x
(We know the opposite side of OBA and we need to find hypotenuse Ob in the triangle OAB. Hence, we need to consider the ration "sin".)
sin 45 = OA
OB
1 = 50
2 x
x = 50* 2 meters
( The distance from the person to Rajender is 50 2 m).
*****************************************
Exercise - 12.1
(1) A tower stands vertically on the ground. From a point which is 15 meter away from the foot of the tower, the angle of elevation of the top of the tower is 45 deg. What is the height of the tower?
sol) Let, the height of the tower be "H" meter
So, AB = H meter
Since tower is vertical to ground ,
So, /_ABC = 90 deg
From the figure, in right angled triangle ABC
Distance between a point and a tower (CB) = 15 mts
Angle of elevation = 45 deg
/_ACB = 45 deg
RTP :- AB = height of the tower
We know the adjacent side of /_ABC and we need to find opposite side AB in the traingle ABC. Hence we need to consider the ratio of "tan"
Now,
tan 45 deg = opposite--
Adjacent
" = AB
CB
************
Tan 45 deg = 1
************
1 = H
15
H = 15 meter. is the height of the tower.
**********************************
(2) A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making 30 deg angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is 6 m. Find the height of the tree before falling down.
sol) Let the height of a tree before broken is "X" meters
Given :- Tree touches the ground by making 30 deg angle
Given :- Distance between foot and top of a tree(AB) = 6 m
RTP :- Height of the tree before falling
From the figure, in right angled triangle OAB
AB = 6 m
/_AOB = 30 deg
We know the opposite side of /_OAB and we need to find adjacent side OB in the triangle OAB. Hence we need to consider the ratio of "tan"
tan 30 deg = AB
OB
1 = 6
_/3 OB
OB = 6 _/3 m
H = 6 _/3 is he height of tree before falling.
*************************************
(3) A contractor wants to set up a slide for the children to play in the park. He wants to set it up at the height of 2m and by making an angle of 30 deg with the ground. What should be the length of the slide?
sol) Given :- Height(h) = 2 m and
RTP :- OA = ?
From the figure, in triangle OAB
OB = 2 m
/_ OAB = 30 deg
sin 30 = OB
OA
1 = 2
2 OA
OA = 4 m
X = 4 m is the length of the slide
*****************************
(4) length of the shadow of a 15 meter high pole us 5 _/3 meters at &'o clock in the morning. Then what is the angle of elevation of the Sun rays with the ground at the time?
sol) Let Angle of elevation of the Sun rays with the ground /_ACB be "x"
Height of the pole = AB = 15 m
Length of the shadow = BC = 5_/3 m
tan x =
= 3
_/3
= _/3 * _/3
-------------- (_/3 *_/3 = 3 )
_/3
x = _/3
Angle of elevation = 60 deg ( _/3 = tan 60)
**************************************
(5) You want to erect a pole of height 10 m with the support of three ropes. Each rope has to make an angle 30 deg with the pole. What should be the length of the rope?
sol) From the figure
In /_\ PQR,
Height of the pole(QR) = 10
Length of rope (PR) = ?
/_PRQ = 30 deg
We Know :- The adjacent side of /_PRQ and we need to find hypotenuse "PR" in the triangle PQR. Hence, we need to consider the ratio "cos"
cos 30 = QR
PR
_/3 = QR
2 PR
_/3 = 10
2 PR
PR = 20 (_/3 = 1.73)
_/3
PR = 11.56 m is the length of the rope
************************************
(6) Suppose you are shooting an arrow from the top of a building at an height of 6m to a target on the ground at an angle of depression of 60 deg. What is the distance between you and the object?
sol) From the figure, In triangle ABC
Height of the building ( OA) = 6 m
/_BAC = 60 deg
Distance between you and object (AC ) = x
We know the opposite side of /_BAC and we need to find hypotenuse "AC" in the triangle ABC. Hence, we need to consider the ratio "sin"
sin 60 = Opposite
Hypotenuse
sin 60 = BC
AC
_/3 = 6
2 AC
AC = 12
_/3
AC = 4 * 3
_/3
AC = 4 *
------------------
AC = 4 _/3 m is the distance between you and building
***********************************************
(7) An electrician wants to repair an electric connection on a pole of height 9m. He needs to reach 1.8 m below the top of the pole to do repair work. What should be the length of the ladder which he should use. When he climbs it an angle of 60 deg with the ground? What will be the distance between foot of the ladder and foot of the pole?
sol) From the figure,,
height of the pole (PB) = 9 cm
AB = 9 - 1.8 = 7.2 cm
/_BCA = 60 deg
Distance between the foot of ladder and foot of pole (BC) = ?
tan 60 = AB
BC
_/3 = 7.2
BC
BC = 7.2
1.73
BC = 4.161 m
**********************************
(8) A boat has to cross a river. It crosses the river by making an angle of 60 deg with the bank of the river due to the stream of the river and travels a distance of 600 m to reach the another side of the river. What is the width of the river?
sol) Let boat starts from "C" and reaches at "A".
/_ACB = 60 deg
AC = 600 m
cos 60 = BC
AC
1 = BC
2 600
BC = 300 m is the width of the river.
****************************************
(9) An observer of height 1.8 m is 13.2 m away from a palm tree. The angle of elevation of the top of the tree from his eyes is 45 deg. What is the height of the palm tree?
sol) From figure OAB
Distance between observer and a tree (AB) = 13.2 m
/_OBA = 45 deg
(WKT the adjacent side of /_OBA and we need to find opposite side OA in the triangle OAB.Hence , we need to consider the ratio "tan" )
tan 45 = Opposite side
Adjacent Side
tan 45 = OA
AB
1 = OA
13.2
OA = 13.2 m
The height of palm tree is OA + AD = 13.2 + 1.8 = 15 m
*************************************************
(10) In the adjacent figure, AC = 6cm, AB = 5 cm and /_BAC = 30 deg. Find the area of the triangle.
sol) In a /_\ABC, draw BD as perpendicular to AC
⧍ BAD is a right angle triangle.
sin 30 = opposite side
Hypotenuse
sin 30 = BD
AB
sin 30 = x
5
sin 30 * 5 = x
1 * 5 = x
2
5 = x
2
2.5 = x
opposite side (x) or Height of /_\ BAD= 2.5 cm
Area of /_\ ABC = 1/2 * 2.5 * 6
=> 7.5 cm
********************************************************************************
*
(*) Solution for Two triangles
Suppose you are observing the top of the palm tree at an angle of elevation 45 deg. The angle of elevation changes to 30 deg when you move 11m away from the tree.
Let us see how we can find height of the tree.
From figure, we have
AB = 11 m
∠DAC = 30 deg
∠DBC = 45 deg
Let the height of the palm tree CD = h meters
and length of BC = x
AC = 11 + x
Case(!): Traingle BDC
Tan 45 = DC
BC
1 = h
x
x = h ----------(1)
Case (!!): triangle ADC
Tan 30 = DC
AC
1 = h
√3 11 + x
h = 11 + x
√3
h = 11 + x
√3 √3
h = 11 + h
√3 √3
h - h = 11
√3 √3
h (√3 - 1 ) = 11
√3 √3
h = 11 meters
(√3 - 1)
NOTE: Total height of the palm tree is
CD + CE
where CE = AF, which is the height of the girl.
*****************************************
Example-6. Two men on either side of a temple of 30 meter height observe its top at the angles of elevation 30 deg and 60 deg respectively. Find the distance between the two men.
sol) Height of the temple BD = 30 deg
Angle of elevation of one person ∠BAD = 30
Angle of elevation of another person ∠BCD = 60
Let the distance between the first person and the temple ,
AD = x and distance between the second person and the temple,
CD = d
From △ BAD
Tan 30 = BD
AB
1 = 30
√3 x
x = 30 √3-----(1)
(*) From △ BCD
Tan 60 = BD
d
√3 = 30
d
d = 30 ----(2)
√3
from (1) and (2) distance between the persons = BC + BA = x + d
= 30 √3 + 30
√3
= 30 *3 + 30
√3
= 90 +30
√3
= 120
√3
= 120 √ 3
√3* √ 3
= 120√ 3
3
= 40√3 meter.
************************************
*****************************************
Example-7. A straight highway leads to the foot of a tower. ramaiah standing at the top of the tower observes a car at an angle of depression 30 deg.
The car is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60 deg.
Find the time taken by the car to reach the foot of the tower from this point.
sol) Let the distance travelled by the car in 6 seconds = AB = x meters
height of the tower CD = h meters
The remaining distance to be travelled by the car BC = d meters
and AC = AB + BC= (x +d) meters
∠ PDA = ∠ DAP = 30
∠PDB = ∠ DBP = 60
Case(!) : Triangle BCD
Tan 60 = CD
BC
√3 = h
d
h = √3d ---------(1)
case(!!) Triangle ACD
Tan 30 = CD
AC
1 = h
√3 (x+d)
h = (x +d) ---------(2)
√3
From (1) and (2), we have
x +d = √3d
√3
x + d = 3d
x = 2d
x = d
2
Time taken to travel 'x' meters = 6 seconds.
Time taken to travel the distance of 'd' meters
i.e,
x meters = 6 seconds.
2 3
= 3 seconds.
****************************************
Exercise- 12.2
(1) A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower , the angle of elevation of the top of tower is 60 deg. from another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30 deg. Find the height of the tower and the width of the road.
sol) Let Height (h) of tower = AB
Given :- From a point on the other side directly opposite the tower, the angle of elevation of the tower is 60 deg
Hence, /_ACB = 60 deg
Also,
Angle of elevation from point "D" to top of the tower = 30 deg
So, /_ADB = 30 deg
Also, DC = 10 m
RTP :- we need to find height of tower "AB" and width of road "BC"
So, /_ABD = 90 deg
(1) In right angle triangle ACB:-
tan C = Side opposite to angle C
Side adjacent to angle C
tan 60 = AB
BC
_/3 = AB
BC
_/3 * BC = AB
AB = _/3BC -------(1)
(2) In a right angle triangle ADB :-
tan D = Side opposite to angle D
Side adjacent to angle D
tan 30 = AB
BD
1 = AB
_/3 BD
BD = AB --------------(2)
_/3
comparing (1) and (2)
_/3BC = BD
_/3
_/3 * _/3BC = BD
3BC = BC + CD
3BC = BC + 10
3BC - BC = 10
2BC = 10
BC = 5
or
x = 5m
Hence, the width of the road is 10 m
From (1)
AB = _/3BC
AB = _/3 * 5
AB = 5_/3 m is the height of the tower
**************************************
(2) A 1.5 m tall boy is looking at the top of a temple which is 30 meter in height from a point at certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30 deg to 60 deg as he walks towards the temple. Find the distance he walked towards the temple.
sol) From figure,
/_DAC = 30 deg and
/_DBC = 60 deg
Since DE = 30 m,
height AF // CE
AF = CE = 1.5 m
CD = DE -- CE
= 30 - 1.5
= 28.5 m
(!) From /_\ DAC
tan 30 = CD
AC
=> 1 = 28.5
_/3 AC
AC = 28.5 * _/3 m {√3 = 1.73 or 2}
AC = 49.36 -----------(1)
(2) From /_\ DBC
tan 60 = DC
BC
=> _/3 = 28.5
BC
=> BC = 28.5
_/3
=> BC = 16.45 ----------------(2)
Distance he walked towards temple = AB
AB = AC -- BC
AB = (1) --- (2)
AB = 49.36 -- 16.43
AB = 32.93 m is the distance he walked towards the temple.
***************************************************
(3) A statue stands on the top of a 2m tall pedestal. From a point on the ground , the angle of elevation of the top of the statue is 60 deg and from the same point, the angle of elevation of the top of the pedestal is 45 deg. Find the height of the statue.
sol) From the figure,
/_DBC = 45 deg and
/_ABC = 60 deg
The height of pedestal (DC) = x m
The height of statue(AD) = 2 m
from /_DBC = 45 deg
tan 45 = DC
BC
1 = DC
BC
BC = DC --------(1)
(2) /_\ ABC = 60 deg
tan 60 = AC
BC
_/3 = AD + DC
BC
_/3 = 2 + DC
DC -----------> ( BC= DC)
DC_/3 = DC + 2
DC√3 - DC = 2
DC (√3 - 1) = 2
DC = 2
(√3 - 1)
DC = 2
(1.73 -1) (√3 = 1.73)
DC = 2
0.73
DC = 200
73
DC = 2.73 m is the height of the pedestal
*********************************
(4) From the top of a 7m building, the angle of elevation of the top of a cell tower is 60 deg and the angle of depressions to its foot is 45 deg. find the height of the tower.
sol) From the figure
Given :- Angle of elevation ( /_EAD) = 60 deg
Angle of depressions ( /_ACD ) = 45 deg
Since AB & CD are parallel
DC = AB = 7m
Also, AD & BC are parallel
So, AD = BC
taking AC as traversal
∠ ACB = ∠ DAC
∠ACB = 45 deg
R.T.P :- Height(CE)=?
(1) From /_ABC
tan 45 = AB
BC
1 = 7
BC
BC = 7 m
Since BC = AD
Also, AD = 7m
(2) From /_EAD
tan 60 = ED
AD
_/3 = ED
7
ED = (7 * _/3) = (7 * 1.73) = 12. 11 m
Height of the tower is :-
=> ED + DC
=> 12.11 + 7
=> 19. 12 m
***************************************************
(5) A wire of length 18m had been tied with electric pole at an angle of elevation 30 deg with the ground. Because it was covering a long distance, it was cut and tied at an angle of elevation 60 with the ground. . How much length of the wire was cut?
sol) From the figure
(1) In a /_\ ABC,
AB = 18
/_ABC = 30 deg
sin 30 = AC
AB
1 = AC
2 AB
1 = AC
2 18
AC = 9 m.
(2) In a /_\ADC
AC = 9 m
/_ADC = 60 deg
sin 60 = AC
AD
√3 = 9
2 AD
AD = 18
_/3
Length of the wire cut = AB -- AD
=> 18 -- 18
_/3
=> 18 -- 18
1.73
=> 18 -- 10.40
=> 7.61 m is the length of the wire had been cut.
******************************************************
(6) The angle of elevation of the top of a building from the foot of the tower is 30 deg and the angle of elevation of the top of the tower from the foot of the building is 60 deg. If the tower is 30 m high, find the height of the building.
sol) From the figure
/_ACB = 60 deg and
/_DBC = 30 deg
Height of tower(AB) = 30 m
Height of building (DC) = x
(1) From /_ACB
tan 60 = AB
BC
_/3 = 30
BC
BC = 10_/3
BC = 10 * 1.73
BC = 17.3 m
(2) From /_DBC
tan 30 = DC
BC
1 = DC
_/3 10_/3
DC = 10m is the height of building
*********************************** *****************
(7) Two poles of equal heights are standing opposite to each other on either side of the road which is 120 feet wide. From a point between them on the road , the angles of elevation of the top of the pole are 60 deg and 30 deg respectively. Find the height of the poles and the distances of the point from the poles.
sol) From the figure
(1) In DPC
tan 30 = CD
CP
1 = CD
_/3 CP
or
CD = CP ------(1)
_/3
(2) In /_\ APB
tan 60 = AB
PB
_/3 = AB
PB
_/3 BP = AB
CD = √3 BP---(2) ( Pole AB = Pole CD)
equate (1) and (2)
we get
CP = √3 BP
√3
CP = 3 BP (√3 * √3 = 3)
Now,
BC = BP + CP
120 = BP + 3BP (CP = 3BP)
120 = 4BP
30 = BP
Now,
CP = BC - BP
CP = 120 - 30
CP = 90 m
From (2)
CD = √3 BP
CD = √3 * 30
CD = 1.73 * 30
CD = 51.9 is the length of the pole
Distance of Point(p) from pole CD = CP = 90m
Distance of Point(p) from pole AB = BP = 30 m
***************************************************
(8) The angles of elevation of the top of a tower from two points at a distance of 4m and 9m . find the height of the tower from the base of the tower and in the same straight line with it are complementary.
sol) From the figure
Given :- Angles are complementary.i.e,
π +π· = 90
π· = ( 90 - π)
(1) In /_\ ABD
tan π = AB
DB
tan π = AB -----------(1)
9
(2) In /_\ ABC
Tan π· = ( AB )
BC
**************
π +π· = 90 deg
π· = (90 -π)
**************
tan (90- π
BC
cot
4
*************
We know
cot
tan
*************
1 = AB
tan π 4
tan π = 4 ----------(2)
AB
equating (1) & (2)
AB = 4
9 AB
AB^2 = 9*4
AB^2 = 36
AB^2 = √36 (6*6 or 6^2 =36)
AB = 6m is the height of the tower.
****************************************************
(9) The angle of elevation of a jet plane from a point A on the ground is 60 deg. After a flight of 15 seconds , the angle of elevation changes to 30 deg. If the jet plane is flying at a constant height of 1500_/3 meter, find the speed of the jet plane ( _/3 = 1.732}
sol) From the figure
/_BAE = 60 and /_CAD = 30
(1) In /_CAD = 30
tan 30 = CD
AD
1 = 1500 _/3
_/3 AD
√3 *√3 = 1500
AD
3 = 1500
AD
AD = 3 * 1500
AD = 4500
(2) In /_BAE = 60
tan 60 = BE
AE
AE
AE = 1500
From figure
BC // ED or BC = ED
ED = AD -- AE
=> 4500 -- 1500
=> 3000m is the distance traveled from point(E) to point(D) in 15 seconds
We have
Speed = distance
Time
=> 3000
15
=> 200 m/s is the speed of the plane
***************************************************
(*) Optional Exercise
1) A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60 deg. After some time, the angle of elevation reduces to 30 deg. Find the distance traveled by the balloon during the interval.
sol) Given :- 1.2 m tall girl spots a balloon at a height of 88.2 m
The height(DE) of the balloon from the eyes of a girl will be
DE = DG - EG
DE = 88.2 - 1.2 = 87 m
Case(!) : In ◿DCE
Tan 30 = DE
CE
1 = 87
√3 CE
CE = 87 √3
Case (!!) : In ◿ABC
Tan 60 = AB
BC
√3 = 87
BC
BC = 87
√3
Multiply and divide by "√3"
BC = 87 √3
√3* √3
BC = 87 √3
3
BC = 29 √3
From the figure
Distance traveled (BE) = CE - BC
=> 87 √3 - 29 √3
=> √3 ( 87 -29)
=> 58√3 m.
*******************************************************
2) The angle of elevation of the top of a tower from the foot of the building is 30 deg and the angle of elevation of the top of the building from the foot of the tower is 60 deg. What is the ratio of heights of tower and building.
sol)
Height of tower(h1) = AB
Height of building(h2) = CD
Distance between tower and building(BC) = x meters
Case(!) : △ ABC,
Tan 30 = AB
BC
1 = h1
√3 x
x = h1 √3 ----------------(1)
Case(!!) : in △BCD
Tan 60 = CD
BC
√3 = h2
x
x = h2 ---------(2)
√3
comparing equation(1) and (2) we get
h1√3 = h2
√3
h1 = 1
h2 √3 *√3
h1 = 1
h2 3
.^., Ratio of height of tower and building
h1 : h2 = 1 : 3
***********************************************
3) The angles of elevation of the top of a lighthouse from 3 boats A,B and C in a straight line of same side of the light house are a, 2a, 3a respectively. If the distance between the boats A and B is "x" meters. Find the height of light house?
sol)The angles of elevation of the top of the lighthouse from the boats
∠PCQ = 3a
∠PBQ = 2a
∠PAQ = a
The distance between the boats A and B = x meters
The height of the lighthouse PQ = ?
Case(!) : ◿ PCQ
tan 3a = PQ
QC
tan 3a = h
QC
QC = h
tan 3a --------(1)
Case(!!) : In◿ PBQ
Tan 2a = PQ
QB
Tan 2a = h
QB
QB = h
Tan 2a ------(2)
Case (!!!) : In ◿ PAQ
Tan a = PQ
QA
Tan a = h
QA
QA = h
Tan a -------(3)
********************
But, from figure
QA = QC + BC + AB
*********************
QC + BC + AB = h
tan a
Given : AB = x mts
Now,
QC + BC + x = h
tan a
******************
from figure
BC = QB - QC
***************
tan a
QB + x = h
tan a
Substitute(2) in above
h + x = h
Tan 2a Tan a
h + x Tan 2a = h
Tan 2a Tan a
cross multiply , we get
Tan a *( h + x Tan 2a) = h Tan 2a
h Tan a + x Tan a Tan 2a = h Tan 2a
h Tan 2a - h Tan a = x Tan a * Tan 2a
h ( Tan 2a - tan a) = x Tan a * tan 2a
h = x * Tan a * Tan 2a
Tan 2a - Tan a is the height of lighthouse
******************************************************
4) Inner part of a cupboard is in the cuboidal shape with its length, breadth and height in the ratio 1: √2 : 1. What is the angle made by the longest stick which can be inserted cupboard with its base inside.
sol) Given :- Inner part of a cupboard is in the cuboidal shape
Length(BC) = 1
Breadth(AB) = √2
Height(AE) = 1
We know that longest stick will be along the body diagonal of the cupboard.
Longest stick in the cupboard= EC
In △ABC
Using Pythagoras Theorem
AC^2 = AB^2 + BC^2
AC^2 = (√2)^2 + (1)^2
AC^2 = 2 +1
AC = √3 ----------(1)
In △ACE
Angle made by the longest stick with the base of the cupboard is given by,
Tan π½ = height of the cupboard
Diagonal of the base of cupboard
Tanπ½ = AE
AC
=> 1
√3 <----(from(1)
Tan π½ = 30 deg
.^., the angle made by the stick with the base is 30 degrees.
******************************************************
5) An iron spherical ball of volume 232848 cm^3 has been melted and converted into a cone with the vertical angle of 120 deg. What are its height and base?
sol) Given : Spherical ball Volume = 232848 cm^3
vertical angle = 120 deg
Let the radius of the base = r
Let the height of the cone = h
Since vertical angle is 120 deg
Height divides this angle into half 60 deg
r = Tan 60
h
r = h√ 3 ----(1)
****************************
We have
Volume of Cone = 1 π *r^2 *h
3
*****************************
Since the iron spherical ball converted into a cone BUT VOLUME is same .
1 * π * r^2 * h = 232848
3
1 * π * (h √3)^2 * h = 232848
3
πΏ*h^3 = 232848
h^3 = 232848
πΏ
h^3 = 232848
(22/7 )
h^3 = 232848 * 7
22
h^3 = 74088
h^3 = 42*42*42 = 42^3
h = 42
So, the height of the cone is 42 cm.
put h = 42 in (1)
we get
r = 42* √3
r = 72.74
Finding Base :
Base = πΉ * r^2
=> πΏ * (72.74)^2
=> 3.14 * (72.74)^2
=> 16614.077 cm^2
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100 cm = 1 m
10000 square cm = 1 square m
**************************
=> 1.6614 m^2
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