Question 1 1 / -0
A ladder, 24 m long, is placed against a wall such that it touches the wall at midway. At the foot of the ladder, the angle of elevation of the midpoint of the wall is 45°. Find the height of the wall.
Solution
Let`s say the height of the wall = h metres.
The ladder is touching the wall at `C`, the height BC is
.
In ΔABC, sin 45° =
h = 24
m
Question 2 1 / -0
The length of the shadow of a pole is
times the height of the pole. Find the angle of elevation of the sun.
Solution
Let the height of the pole be h.
Then, the length of its shadow will be
=
h.
We have to find the angle of elevation of the sun, i.e.
.
As the angle of elevation of the sun is the angle of elevation of the top of the pole from the tip of the shadow, therefore
tan
=
tan
=
tan
=
We know, tan 30° =
So,
= 30°
Question 3 1 / -0
If the shadow of a 5 ft-tall man is 5
ft long, then the angle of elevation of the sun is
Solution
Let the height of the man and the length of his shadow be h and l, respectively.
The angle of elevation of the sun is the angle of elevation of the top of the man's head from the tip of the shadow.
Now, tanθ =
… (1)
Also, tan 30° =
… (2)
On comparing (1) and (2), we get θ = 30°
Hence, the angle of elevation of sun = 30°
Question 4 1 / -0
From two points A and B on the opposite sides of a tower, the angles of elevation to the top of the tower are 45° and 30°, respectively. If the height of the tower is 120 m, then find the distance between A and B, corrected to two decimal places.
Solution
Let the length of the tower be CD.
CD = 120 m
DAC = 45° and
DBC = 30°
In ΔADC,
= tan 45°
= 1
AC = 120 m
In ΔBDC,
= tan 30°
CB = 120
1.732
CB = 207.84 m
Thus, distance between points A and B = AC + CB = 120 + 207.84 = 327.84 m
Question 5 1 / -0
Due to wind, a tree broke from a height and fell on the ground. Its upper top touches the ground at a distance of 6 m from the tree, making an angle of 60° with the ground. What is the height of the tree?
Solution
Let the height of tree be BA + AC and tree broke from point A.
In
ABC,
= tan 60°
AB =
m
And
= sec 60°
AC = 12 m
So, height of the tree = (12 + 6
) m
Question 6 1 / -0
A vertical building and a tower are on the same ground level. From the top of the building, the angle of elevation of the top of the tower is 45° and the angle of depression of the foot of the tower is 60°. Find the height of the tower, if the height of the building is 30 m.
Solution
In the diagram, BC = 30 m = DE,
ABD = 45° and
EBD = 60°.
Let AD = x m and AE = (x + 30) m.
In right-angled ΔABD,
tan 45° =
1 =
BD = x m … (1)
In right-angled ΔEBD,
tan 60° =
=
BD =
m … (2)
From equations (1) and (2), we get
x =
x =
=
m
∴ Height of the tower = AE = (x + 30) m
= (10
+ 30) m
= (10
1.732 + 30) m = 47.32 m
Question 7 1 / -0
The length of the shadow of a vertical pole on level ground increases by 25 metres when the altitude of the sun changes from 60° to 30°. Calculate the height of the pole.
Solution
Let AB be the vertical pole and AD be the shadow when sun's altitude is 60°.
BDA = 60°
Similarly,
BCA = 30°
CD = 25 m (Given)
In
ADB,
tan 60° =
=
AD =
… (1)
In
BAC,
= tan 30°
AB =
AB
=
3AB = 25
+ AB
2AB =
AB = 12.5
≈ 21.65 metres
Question 8 1 / -0
From the top of a building, the angles of depression of two cars standing on the road in a straight line are 30° and 45°, respectively. What is the distance between the cars, if the building is 100 m high?
Solution
In ΔABC,
= tan 45°
AB = BC
BC = 100 m
In ΔABD,
= tan 30°
=
BD = 173.2 m
Distance between the cars = CD = 173.2 - 100 = 73.2 m
Question 9 1 / -0
A flagstaff of height h is mounted on the top of a building. From a point on the ground, the angles of elevation of the foot and the top of the flagstaff are
and
, respectively. If k is the height of the building, then which of the following relations is true?
Solution
Since tan
=
,
So, in
BCD,
tan
=
BC =
…(1)
tan
=
BC =
…(2)
From (1) and (2),
… (3)
Also, cot
=
So, from (3), k cot
= (h + k) cot
is a correct relation.
Question 10 1 / -0
There is a small land mass in the middle of a river with a tree of height 25 m. From two points on each of the opposite banks, the angles of elevation to the top of the tree are 30° and 45°. The two points and a point located at the foot of the tree are in the same straight line, which is perpendicular to both the parallel banks of the river. What is the width of the river?
Solution
In
ABC,
= tan 30°
25
m = BC
BC = 43.3 m
In
ABD,
= tan 45°
BD = 25 m
Width of the river = (43.3 + 25) m = 68.3 m
Question 11 1 / -0
The angles of depression of two houses from the top of a tower are 45° and 60°. One house is directly behind the other. What is the distance between the houses if the height of the tower is 120 m?
Solution
Given, BD = Height of the tower = 120 m
In
ADB,
= tan 45°
= 1
AB = 120 m
In
CDB,
= tan 60°
=
CB =
Now, distance between the houses A and C = AB - CB =
m
= (120 - 69.28) m
= 50.72 m
Question 12 1 / -0
The angle of elevation of the top of an unfinished tower at a point 120 m above its base is 45°. How much must the tower's height be increased so that the angle of elevation becomes 60°?
Solution
Let the height of unfinished tower be h, distance from the base of the tower to the point A be y and the length to be increased be x.
In
ABC
tan 45° = h/y
∴ y = 120 m
Now,
In
ABD:
tan 60° = (120 + x)/120
∴ x = 120(
- 1) m
Question 13 1 / -0
The angle of elevation of the top of a tower from a point on the ground at some distance from its base is 60°. The angle of elevation of the top of the tower from a point 20 m above the same point on the ground is 30°. What is the height of the tower?
Solution
AD is the tower and BC = 20 m
In triangle ACD,
= tan 60°
=
AD = AE + ED
ED = BC = 20 m
=
AE + 20 =
CD . ....,. (1)
In triangle ABE,
= tan 30°
= 1/
Since,
BE = CD
CD =
AE
On putting the value in equation 1, we get
AE + 20 = 3AE
2AE = 20
AE = 10 m
So, AD = 10 + 20 = 30 m
The height of the tower is 30 m.
Question 14 1 / -0
The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of its image in lake is 60°. The height of the cloud is
Solution
Let A be the point of observation and C be the position of the cloud. Let DE be the surface of the lake and C' be the reflection of C in the lake.
CAB = 30°
And
C'AB = 60°
Let BC = h m
In right-angled ΔABC,
= tan 30°
AB =
h … (i)
In right-angled ΔABC',
= tan 60°
(
BC` = BE + EC')
h = 60 m
Height of cloud = 120 m
Question 15 1 / -0
Two flagstaffs stand on a horizontal plane. A and B are two points on the line joining their feet and between them. The angles of elevation of the tops of the flagstaffs as seen from A are 30° and 60° and as seen from B are 60° and 45°. If AB is 30 m, the distance between the flagstaffs (in metres) is
Question 16 1 / -0
The angles of elevation of a plane flying at a constant altitude of 10000 ft are found to be 60° and 30° at an interval of 1 minute. What is the speed of the plane?
Solution
In the figure, BE = CD = 10000 ft
BAE = 60° and
CAD = 30°
In triangle ABE,
In triangle ACD,
Distance travelled in one minute = ED = BC = AC – AB =
Hence, speed =
Question 17 1 / -0
A pole stands vertically on the ground. Two buckets are placed on either side of the pole such that both the buckets and the foot of the pole lie on the same straight line. The angles of elevation of the top of the pole from the two buckets are 45° and 60°, respectively. If the height of the pole is 10 m, find the distance between the two buckets.
Solution
Let B
1 and B
2 be two buckets.
We have to find the distance between the two buckets, i.e. B
1 and
B
2 .
In
ACB
1 ,
Also, tan
m
And in
ACB
2 , tan 45° =
Now, B
1 B
2 = B
1 C + B
2 C
B
1 B
2 =
B
1 B
2 =
Hence, the distance between the two buckets =
m
Question 18 1 / -0
A rocket fired vertically moves according to the relation s = at + b, where s is in kilometres and t in seconds. When it was observed from a point on the ground, which is at a distance of 3 km from the point of projection, it was found that at t = 1 second, the angle of elevation was 30° and at t = 2 seconds, the angle of elevation was 45°. Find the time at which the angle of elevation is 60°.
Solution
In figure, the vertical line gives the position of rocket at different times. At B, the time is 1 second.
In triangle OBA,
OB = a + b and OA = 3
At t = 2 seconds, the position of the rocket is at C.
OC = 2a + b
In triangle OCA,
Solving for b and a,
.
Hence, the expression for the distance becomes
In triangle ODA,
Question 19 1 / -0
A person sitting in an aeroplane, which is flying at a certain height, observes the angles of depression of two consecutive milestones lying at a distance of 1 km on the road to be x and y. Determine the height of the aeroplane above the ground.
Solution
In figure, BC = 1 km
In triangle APB,
In triangle PAC,
Equating the expressions, we get
Question 20 1 / -0
A pole of length 'L', leaning against a wall makes an angle x. When the foot of the pole was moved by a distance of 'b' towards the wall, the top of the pole moves by a distance 'a' upwards and the pole makes an angle y. What is the value of L?
×
Sign in
Profile
Signout
Get latest Exam Updates 
