Exercise 12.1: Some Applications of Trigonometry Q1. A tower stands vertically on the ground. From a point on the ground, 20 m away from the foot of the tower, the angle of elevation of the top the tower is 60°60° . What is the height of the tower? Soln: Given: Distance between point of observation and foot of tower = 20 m = BC Angle of elevation of top of tower = 6O°=06O° = 0 Height of tower H = AB
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Exercise 12.1: Some Applications of Trigonometry · Exercise 12.1: Some Applications of Trigonometry Q1. A tower stands vertically on the ground. From a point on the ground, 20 m
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Exercise 12.1: Some Applications of Trigonometry
Q1. A tower stands vertically on the ground. From a point on the ground, 20 m away
from the foot of the tower, the angle of elevation of the top the tower is 60°60°. What is the height of the tower?
Soln:
Given:
Distance between point of observation and foot of tower = 20 m = BC
Angle of elevation of top of tower = 6O °=06O ° = 0
Height of tower H = AB
Now from fig ABC
AABCAABC is a right angle triangle
1tan0 — oppositeside(AB)Adjacentside(BC) qopposite side(A B ) A d jacen t s ide(B C )
i.e tanC=ABBCtan(7 A BB C
AB = 20 tan60°tan 60°
H = 20><V320 x y / S
= 20V320A/3
.•.heightofthetower=20V3m.\ height o f the tower = 20\/3m
Q2.The angle of elevation of a ladder against a wall is 60°60° and the foot of the
ladder is 9.5 m away from the wall. Find the length of the ladder.
Soln:
Given
A
Distance between foot of the ladder and wall = 9.5 m
C O S 0— adjacentsidehypotenuse COS 0 =adjacent side
cos60°= bcaccos60° =hypotenuse
AC = 2x9.5=19m 2 x 9.5 = 19m
•V . length of the ladder (L) = 19 m
Q3.A ladder is placed along a wall of a house such that its upper end is touching the top the wall. The foot of the ladder is 2 m away from the wall and the ladder is making
an angle of 60°60° with the level of the ground. Determine the height of the wall.
Soln:
A
h
B C2
Distance between foot of the ladder and wall = 2 m =BC
ta n 0 — oppositesideadjacentside tan 0 opposite side tan60°=ABBCtan60° = V 3 = a b2ad jacen t side
AB = 2V32\/3
.•.heightofthewall=2V3m.\ h e i g h t o f t h e w a l l = 2 \/3 m
Q4.An electric pole is 10 m high. A steel wire to top of the pole is affixed at a point on
the ground to keep the pole up right. If the wire makes an angle of 45°45° with the horizontal through the foot of the pole, find the length of the wire.
Soln:
A
Height of the electric pole H = 10 m = AB
Angle made by steel wire with ground (horizontal) 0=45°@ = 45°
Let length of wire = L = AC
If we represent above data in from of a figure then it forms a right angle triangle ABC.
H 2=(90x15)2289i?2 = H2=(90x1517) H 2 = ( ^ y 5) 2 H = 90*1517)
j j __ 9 0 x 1 5 ^
H = 79.41
.•.heightofthekitefromground=79.43mh e i g h t o f t h e k i t e f r o m g r o u n d — 79.43m
Q7.A vertical tower stands on a horizontal place and is surmounted by a vertical flag staff. At a point on the plane 70 meters away from the tower, an observer notices that
the angles of elevation of the meters away from the flag staff are respectively 60°60°
and 45°45° . Find the height of the flag staff and that of the tower
Soln:
A
Given
Vertical tower is surmounted by flag staff.
Distance between tower and observer = 70 m = BC
Angle of elevation of top of tower a=45° a = 45°
Angle of elevation of top of flag staff (3=60°/3 = 60°
Height of flag staff = h = AD
Height of tower = H = AB
If we represent the above data in the figure then it forms right angle trianglesAABCandABCDAABC and A BCD
When 0 0 is angle in right angle triangle we know that
height of broken points from ground = =15(2V3-3)m= 15 (2 \/3 — 3) m
Q9.A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height 5 meters. At a point on the plane, the angles of elevation of the bottom
and the top of the flag staff are espectively 30°30° and 60°60°- Find the height of the tower.
Soln:
A
Height of the flag staff h = 5 m =AP
Angle of elevation of the top of flag staff = 60°60° = acc
Angle of elevation of the bottom of the flagstaff = 60°=(360o = ft
Let height of tower be ‘H’ m = AB
If we represent the above data in forms of figure then it forms a right angle triangle CBD in which ABC is included
Q10.A person observed the angle of elevation of a tower as 30°30°. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of
the top of the tower as 60°60°. Find the height of the tower.
Soln:
A
Given
Angle of elevation of top of tour from first point of elevation (A)a=30°(A) a — 30°
Let the walked 50 m from first point (A) to (B) then AB = 50 m
Angle of elevation from second point (B)(3=60°(.B) /3 = 60°
Now let us represent the given data in form of figure
Then it forms AACDABCD AACD A BCD in which ZC=90 °Zc = 90°
tan(3=ABB Dtan/3 = ta n 4 5 circ= H x+ 20 ta n 45c*rc — ^
x+10 = H
x = H - 10 — (b)
Substitute x = H - 10 in (a)
H - 1 0 = HsqrtS-J^
V3H-10V3=Hv/3-H'-l0v/3 = H H = 10V3V3-1 H = H = io V3x(V3+1)(V3-1)x(V3+1)y3—1
t j 10\ / 3x (-\/3+ l ) D- lO v ^ t v ^ + l )H = (Vs-iMv's+i) H - i o M ^ ) 2 H = --- 2--------
H = 5(3+V3)5 (3 + \/3)
H = 23.66 m
Q12.A parachute is descending vertically and makes angles of elevation of 45°45°
and 60°60° at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of point where he falls on the ground from the just observation point.
Soln:
let the parachute at highest point A. let C and D be points which are 100 m apart on ground where from then CD = 100 m
Angle of elevation from point C = 45°=a45° = a
Angle of elevation from point D = 60°=(360o = [3
Let B be the point just vertically down the parachute
Let us draw figure according to above data then it forms the figure as shown in which ABC and ABD are two triangles
Maximum height of the parachute from the ground
AB = H m
Distance of point where parachute falls to just nearest observation point = x m
If in right angle triangle one of the included angle is 0 0 then
tana=CDADtana = tan30°=Hi50+xtan30° = -tS t-A D 150+a;
150+ x= H V 3 tf\/3 — (a)
tan(3=CDADtan/? = ^ tan60°=Hxtan60° = —J±JJ X
H = X ^ l 3 x y / 3
x = H V3^ — (b)
Substituting (b) in (a)
150 + hV3=HV3-^ - Hy/3
H ( V 3 - iV 3 ) = 1 5 0 i3 ‘ ( A/ 3 - - ^ ) = 1 5 0 H ( 3 - 1 V 3 ) = 1 5 0 JH ' ( ^ ) = 150 H = 150W32
t t 1 5 0 x ^ 3 ^ ~ 2
H = 75V375\/3
H = 129.9 m
•V . height of the tower = 129.9 m
Q16.The angle of elevation of the top of a tower from a point A on the ground is 3030°. Moving a distance of 20 meters towards the foot of the tower to a point B the
angle of elevation increases to 60°60°. Find the height of tower and the distance of the tower from the point A.
Soln:
D
Angle of elevation of top of the tower from point A, a=30°O! = 30°.
Angle of elevation of top of tower from point B, (3=60o/3 = 60°.
Distance between A and B, AB = 20 m
Let height of tower CD = ‘h’ m
Distance between second point B from foot of the tower be ‘x ’ m
If we represent the above data in form of figure then it form a figure as shown with zD=90° / . D = 90°
In right angle triangle, one of included angle is 0@
h(V3-W 3)=20/i ( y $ - X ) = 20 h(3-W3)=20h = 20 h=20*V32h = ^ 1
h = 10V 3 l0v /3
h = 17.32 m
X= 10V 3V 3^
x = 10 m
Height of the tower = 17.32 m
Distance of the tower from point A = (20 + 10) = 30 m
Q17. From the top of a building 15 m high the angle of elevation of the top of tower is
found to be 30°30°. From the bottom of the same building, the angle of elevation of
the top of the tower is found to be 60°60°. Find the height of the tower and the
distance between the building and the tower.
Soln:
Let AB be the building and CD be the tower.
Height of the building is 15 m = h = AB
Angle of elevation of top of the tower from top of the building, a=30°a = 30°.
Angle of elevation of top of the tower from bottom of the building, (3=60°/3 = 60°.
Distance between the tower and the building BD = x
Let height of the tower above building be ‘a’ m
If we represent the above data in form of figure then it form a figure as shown with
zD=90°alsoAX||BD,zAXC=90°ZZ) = 90° also A X \ \ BD, ZAXC = 90°
Here ABDX is a rectangle
BD = DX = ’x’ m AB = XD = h = 15 m
In right triangle if one of the include angle is 0 0
Then tanO—oppositesideadjacentsidetan 0opposite side ad jacen t side
tana=cxAXtana = tan30°=axtan30° = -A X x
x = aV 3a-s /3 — (1)
tan(3=CDBDtan/3 = tan60°=a+i5xtan60° —
xV3=a+15a;\/3 = a + 15 — (2)
Substituting (1) in (2)
xxV3 = aV3a+15 — % = —x-s/3 a+15
a + 15 = aV3(V3)av/3 (>/3)
a + 15 = 3a
2a = 15
a = 15/2 = 7.5 m
x = aV 3a\/3
x = 7.5W 37.5 x V 3
x = 12.99 m
Height of the tower above ground = h + a
= 15 + 7.5
= 22.5 m
Distance between tower and building = 12.99 m
Q18.On a horizontal plane there is a vertical tower with a flag pole on the top of the tower. At a point 9 m away from the foot of the tower the angle of elevation of the topand bottom of the flag pole are 60°60° and 30°30° respectively. Find the height of the tower and the flag pole mounted on it.
Soln:
D
let AB be the tower and BC be the flagstaff on the tower
Distance of the point of observation from foot of the tower BD = 9 m
Angle of elevation of top of flagstaff, a=60°a = 60°
Angle of elevation of top of flagstaff, (3=30o/3 = 30°
Let height of the tower = ‘x’ = AB
Height of the pole = ‘y’ = BC
If the above data is represented in form of figure a shown with zA=90°Z.A = 90°
In right angle triangle, one of included angle is 0@
Then tanO—oppositesideadjacentsidetan 0opposite side adjacent side
tana=ACADtana = tan60°=x+y9tan60° = ^ -
x + y= 9V39\/3 — (a)
tan(3=acad tan/? = tan30°=x9tan30° = f
x = 3V33\/3
x = 5.196 m — (b)
Substituting (b) in (a)
y = 9V3-3V3973 - 3\/3
y = 6V3673
y = 10.392 m
•V. Height of the tower x = 5.196 m
Height of the tower pole = 10.392 m
Q19.A tree breaks due to storm and the broken part bends so that the top of the treetouches the ground making an angle of 30°30° with the ground. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Soln:
A
B
h
C8
Let the initial height be AB
Let us assumed that the tree is broken at point C
Angle made by the broken part CB’ with ground is 3O°=03O° = 0
Distance between foot of tree to point where it touches the ground B’A = 8 m
Height of the tree = h = AC + CB’ = AC + CB
The above information is represent in the form of figure as shown
C O S 0 —AdjacentsideHypotenuseCOS 0
^ 2=8C B '4 =
Adjacent side Hypotenuse cos30°=A B C B C O S 3 0 ° A H
C H
CB’ = 16V3^| V3
tan0—oppositesideadjacentsidetan0 = °PPosite tan30°— C A A B 'tan 30° = -7^7adjacent side A H
^ = CMSj , = OA
CA =
Height of the tree = CB’ + CA = 16V3 + 8V3-J! + 8V3
h = 24V3 -7=
h= 8'l3rr\8y/3m
Q20.From a point P on the ground the angle of elevation of a 10 m tall building is 30° 3 0 ° . A flag is hoisted at the top of the building and the angle of elevation of the flag
staff from P is 45°45°. Find the length of the flag staff and the distance of the building from the point P.
Soln:
let AB be the tower and BD be the flag staff
Angle of elevation of top of the building from P a=30°Q! = 30°
AB = height of the tower = 10 m
Angle of elevation of top of flag staff from P (3=45°/3 = 45°
Let height of the flagstaff BD = ‘a’ m
The above information is represented in the form of figure as shown
tanO = oppositesideadjacentsidetan 0 = °PP°site tana=ABAPtana =adjacent side AP
tan30°=iOAPtan30° = AP
A P = 1(^31711(^/3 m
A P = 17.32 m
tan(3=ADAPtan/? = ^ tan45°=io+aAPtan45° =
10 + a = AP
a = 17.32 -1 0 m
a = 7.32 m
Height of the flag staff ‘a’ = 7.32 m
Distance between P and foot of the tower = 17.32 m
Q21 .A 1.6 m tall boy stands at a distance of 3.2 m from a lamp post and casts a shadow of 4.8 m on the ground. Find the height of the lamp post by using
(i) trigonometric ratio
(ii) properties of similar triangles
Soln:
4,8 E 3.2
Let AC be the lamp post of height ‘h’
We assume that ED = 1.6 m, BE = 4.8 m and EC = 3.2
We have to find the height of the lamp post
Now we have to find the height of the lamp post using similar triangles
Since triangle BDE and triangle ABC are similar
ACBC = E D B E ^ - h4.8+3.2 = 1.64.8 4 g 2 = h=83ITl/i = f 771
Again we have to find the height of lamp post using trigonometric ratio
h*V3=h+1.6h x y/3 = h + 1.6 h( sqrt3-1 )=1.6h{ sqrtS — 1) = 1.6
h= 1 ^ - 1 x ^ 1
h = 0.5( sqrt3+1 )0.5 ( sqrtS + 1)
Height of the pedestal = 0.6( sqrt3+1)0.6 ( sqrtS + 1)
Q27.A T.V. tower stands vertically on a bank of a river of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is
60°60°. From a point 20 m away this point on the same bank, the angle of elevation of
the top of the tower is 30°30°. Find the height of the tower and the width of the river.
Soln:
A
h
B
let AB be the T.V tower of height ‘h’ m on the bank of river and ‘D’ be the point on the
opposite side of the river. An angle of elevation at the top of the tower is 30°30°
Let AB = h and BC = x
Here we have to find height and width of the river.
The above data is represented in the form of figure as shown.
In AACBAAC5
tan60°=ABBCtan60° = V3=hx\/3 = - ^l3x=hy/3x = hx= hV3cc = A=.d C x y 3
Q31 .From a point on a bridge across a river the angle of depression of the banks on
opposite side of the river are 3 0 ° 3 0 ° and 4 5 ° 4 5 ° respectively. If the bridge is at the height of 30m from the banks, find the width of the river.
Soln:
A
Height of the bridge = 30m [AB]
Angle of depression of bank 1 i.e, a=30°Bia = 3 0 ° B i
Angle of depression of bank 2 i.e, (3=30°B2/3 = 30°B 2
Given banks are on opposite sides
Distance between banks B-|B2=BB-|+BB2.Bi .B2 = BB\ + BB 2
The above data is represented in the form of figure as shown
In right angle triangle, if one of the included angle is 0 0
B1 B2=BBi +BB2B 1B 2 = B B 1 + BB2 3(^3+3030^/3 + 30 30(V3+1)30 (y/3 + l)
Distance between banks = 30(V3+1)m30 (\/3 + l ) m
Q33.A man sitting at a height of 20 m on a tall tree on a small island in the middle of a river observes two poles directly opposite to each other on the two banks of the river and in line with foot of tree. If the angles of depression of the feet of the poles from a
point at which the man is sitting on the tree on either side of the river are 60°60° and
30°30° respectively. Find the width of the river.
Height of the tree AB = 20m
Angle of depression of the pole 1 feet a=60°Bia = 60°Bi
Angle of depression of the pole 2 feet (3=30oB-|/? = 30oi?i
B-iC-i-Bi C i be one pole and B2C2 B 2 C 2 be the other pole
Given poles are on opposite sides.
Width of the river = B 1B2.B1.B2
= B1B+BB2B1B + BB2
Soln:
A
The above data is represented in the form of figure as shown
In right angle triangle, if one of the included angle is 0 0
tan a= A B B iB tan o : = ta n 6 0 °= 2 0 B iB ta n 6 0 ° = B 1B = 2 oV3jB 1jB = -^=tilts tiiti A /3
tanP=ABBB2 t a n p — tan30°=20B2Btan30o — B2B=2(h/3mB 2 B — 2 0 y / 3 m
B i B2=B i B+BB2B i B 2 = B \ B + B B 2 2oV3+20V3-^= -T 20 \/3
2 0 1 +3V3 = 8(W3 ^ = - —
Width of the river = 8 0 V 3 lT I-^ ra
Q34.A vertical tower stands on a horizontal plane and is surmounted by a flag staff of height 7m. From a point on the plane, the angle of elevation of the bottom of flag staff
is 30°30° and that of the top of the flag staff is 45°45°. Find the height of the tower.
Soln:
Given
Height of the flag staff = 7m = BC
Let height of the tower = ‘h’ m = AB
Angle of elevation of top of the bottom of the flagstaff a=30°a = 30°
Angle of elevation of top of the top of the flagstaff (3=45°/3 = 45°
Points of desecration be ‘p’
The above data is represented in the form of figure as shown
ta n a = A B B C ta n a — 4 5 t a n 4 5 ° = h a ta n 4 5 ° — -B C a
h = a — (1)
In A A D B A A O B
tan(3=A B (2x+a)B C tan/3 = 2xf^BC ta n 3 0 °= h 2 x + a ta n 3 0 ° = ^
2x+a = hV3/i\/3 — (2)
Substituting (1) in (2)
2x+h=hV32z + h = h y / 3 h(V3-1)=2x/i ( y / 3 - l ) = 2 x
h=2x,5-1x1gH 15+1 - ^ r x ^ l
h = 2x(V3+i) 2 2a;( 3+12
h = xV3+1)a;\/3 + l )
Height of the tower = x(V3+1 )ma:( v 3 + 1 )m
Q36.A tree breaks due to storm and the broken part bends so that the top of the tree
touches the ground making an angle of 30°30° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 meters. Find the height of the tree.
Soln:
Let AB be the height of the tree and it is broken at point C and top touches ground at B’
Angle made by the top a=30°a = 30°
Distance from foot of tree from point where it touches ground = 10 m
The above data is represented in form of figure as shown
Height of the tree = AB = AC + CB
= AC + CB’
In right angle triangle one of the included angle is 0 @
Sin0=OppositesideHypotenusesin 0 = s id e sin60°= ABBCsin60° = - 577 V32 = h215Hypotenuse B C
V 3 __ h2 ~ 215
h = 2 l W S 2 S | £
h = 107.5V3ml07.5-\/3 m
Height of the balloon from ground = 107.5V3ml07.5'\/3?7i
Q38. Two men on either side of the cliff 80m high observe the angles of elevation of the top of the cliff to be 300 and 600 respectively. Find the distance between the two men.
Soln:
A
Height of cliff = 80m = AB
Angle of elevation from Man 1, a=30°[Mi]a = 30° [M i]
Angle of elevation from Man 2, (3=60°[M2]/3 = 60° [M2]
Distance between two men = M i M 2= B M i + B M 2M i M 2 — BM \ + B M 2
The above information is represented in form of figure as shown
In right angle triangle one of the included angle is 0@
Q44.The angles of depression of two ships from the top of a light house and on the
same side of it are found to be 45°45° and 30°30° respectively. If the ships are 200 m apart, find the height of the light house.
Soln:
3000
D
Height of the light house AB = ‘h’ meters
Let S -ia n d S 2 5 i and S2 be ships
Distance between ships S-|S2=200m5i52 = 200m
Angle of depression of S i (a=30°)5i (a = 30°)
Angle of depression of S 2(3= 45°)< S ,2 = 4 5 ° )
The above data is represented in form of form of figure as shown
In AABS2A A B 52
tan(3=ABBS2t a n /3 = ta n 4 5 °= h B S 2t a n 4 5 0 =
BS2=hB S2 = h — (1 )
In A A B S iA A B 5 i
tana=ABBSitana = tan30°=hBSitan30° =
B S i= h V 3 5 S i = hy/ 3 — (2 )
Subtracting (1) from (2)
BSi-BS2=h(sqrt3-1)S5i--B52 = h ( s q r t 3 - l ) 200=h(sqrt3-1)
200 = h ( s q r t S - l ) h=100(sqrt3+1)meters/i = 100 ( s q r t S + 1) m e t e r s
Height of the light house = 273.2 meters
Q45.The angles of elevation from the top of a tower from two points at distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m.
Soln:
Height of the tower AB = ‘h’ meters
Let point C be 4 meters from B.
Angle of elevation is Q a given point D is 9 meters from B, Angle of elevation is (3/3
Given d a , (3/3 are complementary, d a + (3=90°=>=;(3=90oa/3 = 90° = > = ; /3 = 90° a
Required to prove that h = 6 meters
The above data is represented in the form of figure as shown
In AABC A A B C
tana=ABBCtana = ^ tana=h4tana =
h = 4tana4tana — (1)
In A A B D A A B D
tan(3=ABBDtan/3 = tan(90-a)=h9tan(90 — a ) = ■|
h = 9tana9tana — (2)
Multiply (1) and (2)
h x h = 4 ta n ax 9co ta /i x h = 4 tan a x 9 cot a h2=36(tanacota) h 2 = 36 (tan a cot a) h=V36=6m/i = y/S6 = 6m
Height of the tower = 6 meters
Q46.From the top of a 50 m high tower, the angles of depression of the top and bottom
of a pole are observed to be 45°45° and 60°60° respectively. Find the height of the pole.
Soln:
AB = height of the tower = 50m.
CD = height of the pole
Angle of depression of top of building a=45°a = 45°
Angle of depression of bottom of building (3=60°/3 = 60°
The above data is represented in the form of figure as shown
50(V 3-1 )V3 = 503 (3“ V3) 5 0 ( V 3 - 1 ) 50= f ( 3 - V 3 )
Height of the building (pole) = 503(3-V3)f (3 - V3)
Distance between the pole and tower = 50V3 m =my 3
Q47.The horizontal distance between two trees of different heights is 60 m. the angles of depression of the top of the first tree when seen from the top of the second tree is
45°45°. If the height of the second tree is 80 m, find the height of the first tree.
Soln:
c
Distance between the trees = 60 m [BD]
Height of second tree = 80 m [CD]
Let height of the first tree = ‘h’ m [AB]
Angle of depression from second tree top to first tree top a=45°a = 45°
The above data is represented in form of figure as shown
(iii) The difference between the heights of the building and the lamp post.
Soln:
Q53.From the top of a building AB = 60m, the angles of depression of the top and
bottom of a vertical lamp post CD are observed to be 30°30° and 45°45° respectively.Find
Height of building AB = 60m
Height of lamp post CD = ‘h’ m
Angle of depression of top of lamp post from top of building a=30circa = 30circ
Angle of depression of bottom of lamp post from top of building (3=60circ/3 =
The above data is represented in form of figure
Draw DXlABD X JL AB, CX = AC, CD = AX
In ABDXA B D X
Then tana— OppositesideAdjacentside — BXDXtan a Opposite side Adjacent side
B XD X
tan30°=60- cdac tan 30° = 60 9J* iV3=60-hAC^= =A C Vo A c
AC = (60—h)V3m(60-/i) s/Sm — (1)
In ABCAA BCA
tan (3=abac tan/3 = ^ tan60°=60Actan60° =
AC = 6oV3=20V3m-^| = 20\/3 m — (2)
From (1) and (2)
(60-h)V3=20V3(60- h) V3 = 20^/3
60 - h = 20
h = 40m
Height of the lamp post = 40m
Distance between lamp post building AC = 20V3m20\/3m
Difference between heights of building and lamp post BX = 60 - h => 60 - 40 = 20m.
Q54.Two boats approach a light house in mid sea from opposite directions. The
angles of elevation of the top of the light house from two boats are 30°30° and 45° 45° respectively. If the distance between the two boats is 100m, find the height of the light house.
Soln:
A
let B<\Bi be boat 1 and B2B2 be boat 2
Height of light house = ‘h’ m = AB
Distance between BiB2=100m.Bi.B2 = 100m
Angle of elevation of A from B-ia=30circBi a = 30CTrc
Angle of elevation of B from B2(3=45circ.B2 P = 45CTrc
The above data is represented in the form of figure as shown
Here In AABB-i A A BB i
tan30°—OppositesideAdjacentside —ABBBitan30° = ide ~
BiB=ABV3=hV3BiB = AByJ3 = hy/Z — (1)
In AABB2AARB2
tan45°=ABBB2tan450 =
h = BB_{2} — (2)
Adding (1) and (2)
B1B2=h(V3+1)JB i5 2 = h (y /3 + 1)
h ■
h= 100V3+1 XV3-1V3-1 -125- x ^ z l> /3 + l a/ 3 - 1
h= 100( 3-1)2 = 50(V3-1)loo( 11 = 50 (v/3 - 1)
Height of the light house = 50(V3-1 )50 (\/3 — l)
Q55.The angle of elevation of the top of a hill at the foot of a tower is 60°60° and the
angle of elevation of the top of the tower from the foot of the hill is 30°30°. If tower is 50 m high, what is the height of the hill?
Soln:
Height of tower AB = 50m
Height of hill CD = ‘h’ m.
Angle of elevation of top of the hill from foot of the tower = a=60°Q! = 60°
Angle of elevation of top of the tower from foot of the hill = (3=30°/3 = 30°
The above data is represented in the form of figure as shown
Station 1 should send its team and they have to travel 7.32 km
Q57.A man on the deck of a ship is 10 m above the water level. He observes that the
angle of elevation of the top of a cliff is 45°45° and the angle of depression of thebase is 30°30° . Calculate the distance of the cliff from the ship and the height of the cliff.
Soln:
Height of the ship from water level = 10m = AB
Angle of elevation of top of the cliff a=45°a = 45°
Angle of elevation of bottom of the cliff (3=30°/3 = 30°
Height of the cliff CD = ‘h’ m
Distance of the ship from foot of the tower cliff.
Height of cliff above ship be ‘a’ m
Then height of cliff = DX + XC = (10 + a) m
The above data is represented in form of figure as shown
In right angle triangle, if one of the included angle is 0@
Then tan0=OppositesideAdjacentsidetan@ = ®PP°site sideAdjacent side
tan45°= cxAxtan 45° = 1=aAxl =
AX = ‘a’m
tan30°=XDAXtan30° = 4 ? i V3 = ioax4= = 4 4A X s/3 A X
AX = 10V3l0\/3
.•.a=10V3m.\ a = lOy/Sm
Height of the cliff = 10+10V3m=10(V3+1 )10 + 10 \/3m — 10 ( \ /3 + l )
Q59.There are two temples one on each bank of a river opposite to one other. One temple is 50m high. Form the top of this temple, the angles of depression of the top
and foot of the other temple are 30°30° and 60°60° respectively. Find the width of the river and the height of the other temple.
Soln:
Height of the temple 1 (AB) = 50m
Angle of depression of top of temple 2, a=30°a = 30°
Angle of depression of bottom of temple 2, (3=60o/3 = 60°
Height of the temple 2 (CD) = ‘h’ m
Width of the river = BD = ‘x’ m
The above data is represented in from of figure as shown
Q62.A tree standing on a horizontal plane is leaning towards east. At two points situated at distance a and b exactly due west on it, the angles of elevation of the top
are respectively Ota and P/3. Prove that the height of the top from the ground is
(b-a)tanatanptana-tanp(b—a) tan a tan /J
tan a-tan /?
Soln:
AB be the tree leaning east
From distance ‘a’ m from tree, angle of elevation be a a at point P.
From distance ‘b’ m from tree, angle of elevation be (3/3 at point Q.
The above data is represented in the form of figure as shown
tan 15°=aqay tan 15° = 48 0.268= hAY 0.268 = 48A Y A Y
A Y = h0.268 A Y - ----(1)
tan45°=AB'AYtan45° = 4 8 tan45°=AP+PQ'AYtan45° = AP+Z QA Y A Y
AY = x + (h + x)
AY = h + 2x — (2)
From (1) and (2)
h0.268 h+2x q 2gg h -\ -2 x
3.131h-h = 2x2500
h = 1830.8312
Height of the cloud above lake = h + x
= 1830.8312 + 2500
= 4330.8312 m
Q64.lf the angle of elevation of a cloud from a point h meters above a lake is act: and
the angle of depression of its reflection in the lake be (3/?, prove that the distance ofthe cloud from the point of observation is 2hsecatanp-tana 2/t sec a—tan p—tan a
let x be point ‘h’ meters above lake
Angle of elevation of cloud from x = act
Angle of depression of cloud reflection in lake = (3/3
Height of the cloud from lake = PQ
PQ’ be the reflection then PQ’ = PQ
Draw XA + PQ, AQ = ‘x’ m, AP = ‘h’ m
Distance of cloud from point of observation is XQ.
Soln:
Q
Q'
The above data is represented in form of figure as shown
In AAQXA AQX
tana=AQAXtan a = ^
tana=xAXtana = — (a)
In AAXQAAXQ'
tan3=AQAXtan ft = 4%
tan(3=h+x+hAXtan/3 = h+ h — (2)
Subtracting (1) from (2)
tan(3-tana= 2hAxtan j3- tan a AX- 2htan|3-tana A X 2htan j8-tan a
In AAQXA AQX
cosa=AXXQcosa = ^
QX = AXsecaAX sec a
X Q = 2hsecatanp-tana Z h s e c atan p-tan a
distance of cloud from point of observation = 2hsecatan|3-tana 2 h sec a tan /3-tan a
Q65.From an airplane vertically above a straight horizontal road, the angle of depression of two consecutive mile stones on opposite side of the airplane are
observed to be act and 3/3. Show that the height in miles of airplane above the road is. . tan a tan Bgiven by tanatanptana+tanp tan a_|_tan ■
Soln:
-V Q >-
let PQ be the height of airplane from ground x and y be two mile stones on opposite side of the airplane xy = 1 mile
Angle of depression of x from p = da
Angle of depression of y from p = (3/3
The above data is represented in the form of figure as shown
x , „ , tana.tan/?PQ = tana.tanBtanB+tana -— ^—-----tan p+tan a
Height of airplane = tana.tanptanp+tana tan a.tan /3 tan /3+tan a miles.
Q66.PQ is a post given height ‘a’ m and AB is a tower at same distance. If act and P/3 are the angles of elevation of B, the top of the tower, at P and Q respectively. Find the height of the tower and its distance from the post.
Soln:
PQ is part height = ‘a’ m AB is tower height
Angle of elevation of B from P = act
Angle of elevation of B from Q = (3/3
The above data is represented in gorm of figure as shown
In right angle triangle if one of the included angle is 0@
QX = ABtana = atana-tanp ,AB = -----— —tan a tan a —tan p
Height of tower = atana-tanp ------— wtan a —tan p
Distance between post and tower = atanatana-tanp atanata n a —tan p
Q67.A ladder rest against a wall at an angle d a to the horizontal. Its foot is pulled away from the wall through a distance a, so that it slides a distance b down the wall making an angle (3/3 with the horizontal. Show that ab=cosa-cospsinp-sina
a cos a —cos /3b sin /?—sin a
Soln:
let AB be the ladder initially at an inclination a a to ground
When its foot is pulled through distance ‘a’ let BB’ = ‘a’ m and AA’= ‘b’ m
New angle of elevation from B’ = B the above data is represented in form of figure as shown
Let APIgroundB P ,AB=A B A P ± groundB 'P , A B = A 'B '
A’P = x BP = y
In A A B P A A R P
sina=APABsino: = 4§AB
sina=x+bABsina = ^ — (1)
cosa= bpab cos a = 4 1AB
C0Sa=yABCOSO! = 4 s -----(2)
In AA'B'P A A 'B 'P
sin(3=A'PAB'sin/3 =A B
sin(3=xABsin/3 = -g j — (3)
COS(B= B'PA'B'COS /3 —A B
COS(3= y+aABCOS/3 = 4 4 — (4)AB
Subtracting (3) from (1)
sina—sin(3=bABsina:-sin/? =
Subtracting (4) from (2)
COS(3—COSQ= aAB COS/?— COSO: ab — cosa-cospsinp-sina ^
Q68.A tower subtends an angle Cla at a point A in the plane of its base and the angle
if depression of the foot of the tower at a point b meters just above A is P/3. Prove that the height of the tower is btanacotp&tan a cot f3.
Soln:
p
let height of the tower be ‘h’ m = PQ
Angle of elevation at point A on ground = aa
Let B be the point ‘b’ m above the A
Angle of depression of foot of the tower from B = |3/3
The above data is represented in form of figure as shown
Let B X lP ,AB=A'B 'B X ± P , A B = A 'B '
In A P B X A P B X
tana=PQBXtana — — (1)
In A Q B X A Q B X
tan(3=QXBXtan/3 = ^ — (2)
Dividing (1) by (2)
. ______ _ tan a PQta n o ta n p -P Q Q X -^ =
PQ = QX.tanotanp=btana.cotp^^ = fttana. cot /9tan p
Q69.An observer, 1.5 m tall, is 28.5 m away from a tower 30 m high. Determine the angle of elevation of the top of the tower from his eye.
Soln:
A
Height of observer = AB = 1.5 m
Height of tower = PQ = 30 m
Height of tower above the observer eye = 30 - 1.5
QX = 28.5m
Distance between tower and observer XB = 28.5m
0 0 be angle of elevation of tower top from eye
The above data is represented in the form of figure
tan0=O ppositesideAdjacentsidetan 0 = Opposite side. ta n 0 = Q X B X = 28.528.5 = 1Adjacent side
tan0 = = H I = 1 O=tarr1(1)=45°0 = t a n ^ l) = 45°
Angle of elevation = 45°45°
Q70. A carpenter makes stools for electricians with a square top of side 0.5 m and at a height of 1.5 m above the ground. Also each leg is inclined at an angle of 60°60° to the ground. Find the length of each leg and also the lengths of two steps to be put at equal distance.
Soln:
A B
let AB be the height of stool = 1.5 m
Let P and Q be equal distance then AP = 0.5. AQ = 1 m
The above data is represented in form of figure as shown
BC = length of leg
sin60°=ABBCsin60o — ^ V32 = 1.5BC^ —
BC = 1 .5 *2B C ^^
Draw P X ±A B ,Q Z ±A B ,X Y ±C A ,Z W ±C AP X ± A B , Q Z _L A B , X Y _L C A , Z W _L C A
sin60°=XYXcsin60° = ^
XC = 0.5BCx V 3 -^ X y/s
a/ 3 4 X 8 3 # X | 2V3- 7 =4 3 ^ / 3
XC = 1.1077m
sin60°=zwczsin60° = ^
CZ — 1 sqrt321
sqrtS2
c z = 2 V 5 ^
CZ= 1.654m
Q71.A boy is standing on the ground and flying a kite with 100m of string at an
elevation of 30°30°. Another boy is standing on the roof of a 10m high building and is
flying his kite at an elevation of 45°45°. Both the boys are on opposite sides of boththe kites. Find the length of the string that the second boy must have so that the kitesmeet.
Soln:
A
For boy 1
Length of the string AB = 100 m
Angle made by string with ground = a=30oo: = 30°
For boy 2
Height of the building CD = 10m
Angle made by string with building top = (3=45°/3 = 45°
Length of the kite thread of boy 2 if both the kites meet must be ‘DB’
The above data is represented in form of figure as shown
Draw BX±AC ,Y D l B C B X _L A C , Y D _L B C
In A A B X A A B X
tan30°= BXAXtan 30° = 4$ sin30°=BXABsin30° = 12 = bxioo-A X A B 2
BX = 50m
BY = BX - XY = 50 - 10 = 40m
In A B Y D A S T O
sin45°=BYBDsin45° = iV2=40bd- ^ =
BD = 40V240\/2
Length of thread or string of boy 2 = 40V240\/2
B X100
Q72. From the top of a light house, the angles of depression of two ships on the
opposite sides of it are observed to be d a and (3/3. If the height of the light house be h meters and the line joining the ships passes through the foot of the light house, show
. . . . . . ft(tan a+tan 0)that the distance h(tana+tanB)tanatanB — ------ -— -—tan a tan p
Soln:
Height of light house = ‘h’ meters = AB
S ia n d S 2 S i a n d S 2 be two ships on opposite sides of the light house
Angle of depression of S-|5i from top of light house = act
Angle of depression of S 2S 2 from top of light house = (3/3
Required to prove that
. . . . . ft(tan a+tan/3)Distance between ships = h(tana+tanB)tanatanB —1r----- -— ^ meterstan a tan p
The above data is represented in the form of figure as shown
In A A B S 1A A B S 1
t a n a = OppositesideAdjacentside = A B S iB tano: = ^ ^ a c e rL t side =
Si B= htana S \ B — —^------(1)
In AABS 2 A A B S 2
tan3=ABS2B tan^= ^
S2B=htanpS2B = d b — (2)
Adding (1) and (2)
=> S-| B+S2— htana + htan(3 S \ B + S 2 = ^ ^
=> S iS 2-h{ltana + 1tan3}5i<S2 = h { ^
fo(tan a+tan B) => h(tana+tanp)tanatan(3 — ------ -— -—tan a tan p
, . /i(tan a+tan 0)Distance between ships = h(tana+tanp)tanatanp — ---- -— -a— meterstan ol tan p