ICSE Class 10

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Chapter Test

Question 1.
An electric pole is 10 metres high. If its shadow is 10√3 metres in length, find the elevation of the sun.
Solution:
Let AB be the pole and
OB is its shadow.

Question 2.
The angle of elevation of the top of a tower from a point on the ground and at a distance of 150 m from its foot is 30°. Find the height of the tower correct to one place of decimal
Solution:
Let BC be the tower and
A is the point on the ground such that
∠A= 30° and AC = 150 m

Question 3.
A ladder is placed against a wall such that it just reaches the top of the wall. The foot of the ladder is 1.5 metres away from the wall and the ladder is inclined at an angle of 60° with the ground. Find the height of the wall.
Solution:
Let AB be the wall and AC be the ladder
whose foot C is 1.5 m away from B
Let AB = x m and angle of inclination is 60°

Question 4.
What is the angle of elevation of the sun when the length of the shadow of a vertical pole is equal to its height.
Solution:
Let AB be the pole and CB be its shadow
and θ is the angle of elevation of the sun.
Let AB = x m, then BC = x m

Question 5.
A river is 60 m wide. A tree of unknown height is on one bank. The angle of elevation of the top of the tree from the point exactly opposite to the foot of the tree on the other bank is 30°. Find the height of the tree.
Solution:
Let AB be the tree and BC is the width of the river
and C is the point exactly opposite to B on the other bank
and angle of elevation is 30°.

Question 6.
From a point P on level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, how far is P from the foot of the tower ?
Solution:
Let AB be the tower and P is at a distance of x m from B, the foot of the tower.
While the height of the tower AB = 100 m
and angle of elevation = 30°

Question 7.
From the top of a cliff 92 m high, the angle of depression of a buoy is 20°. Calculate to the nearest metre, the distance of the buoy from the foot of the cliff. (2005)
Solution:
Let AB be cliff whose height is 92 m
and C is buoy making depression angle of 20°.

Question 8.
A boy is flying a kite with a string of length 100 m. If the string is tight and the angle of elevation of the kite is 26°32′, find the height of the kite correct to one decimal place, (ignore the height of the boy).
Solution:
Let AB be the height of the kite A and AC is the string
and angle of elevation of the kite is 26°32′

Question 9.
An electric pole is 10 m high A steel wire tied to the top of the pole is affixed at a point on the ground to keep the pole upright. If the wire makes an angle of 45° with the horizontal through the foot of the pole, find the length of the wire.
Solution:
Let AB be the pole and AC be the wire
which makes an angle of 45° with the ground.
Height of the pole AB = 10 m
and let the length of wire AC = x m

Question 10.
A bridge across a river makes an angle of 45° with the river bank. If the length of the bridge across the river is 200 metres, what is the breadth of the river.

Solution:
Let AB be the width of river = xm
Length of the bridge AC = 200 m
and angle with the river bank = 45°
sin θ = \(\\ \frac { AB }{ AC } \)
⇒ sin 45° = \(\\ \frac { x }{ 200 } \)

Question 11.
A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ? (2001)
Solution:
Let AB be the tower and
let a man C stands at a distance from the foot of the tower = x m
and cos θ = 0.53

Question 12.
The upper part of a tree broken by wind, falls to the ground without being detached. The top of the broken part touches the ground at an angle of 38°30′ at a point 6 m from the foot of the tree. Calculate.
(i) the height at which the tree is broken.
(ii) the original height of the tree correct to two decimal places.
Solution:
Let TR be the total height of the tree
and TP is the broken part which touches the ground
at the distance of 6 m from the foot of the tree
making an angle of 38°30′ with the ground.
Let PR = x and TR = x + y
PQ = PT = y
In right ∆PQR


Height of the tree = 4.7724 + 7.6665 = 12.4389 = 12.44 m
and height of the tree at which it is broken = 4.77 m

Question 13.
An observer 1.5 m tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Solution:
In the figure, AB is tower and CD is an observer.
θ is the angle of observation from

Question 14.
From a boat 300 metres away from a vertical cliff, the angles of elevation of the top and the foot of a vertical concrete pillar at the edge of the cliff are 55°40′ and 54°20′ respectively. Find the height of the pillar correct to the nearest metre.

Solution:
Let CB be the cliff and AC be the pillar
and D be the boat which is 300 m away from
the foot of the cliff i.e. BD = 300 m.
Angles of elevation of the top and foot of the pillar
are 55°40′ and 54°20′ respectively.
Let CB = x and AC = y
In right ∆CBD,

Question 15.
From a point P on the ground, the angle of elevation of the top of a 10 m tall building and a helicopter hovering over the top of the building are 30° and 60° respectively. Find the height of the helicopter above the ground.
Solution:
let AB be the building and H is the helicopter hovering over it.
P is a point on the ground,
the angle of elevation of the top of building and helicopter are 30° and 60°

Question 16.
An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at the instant.
Solution:
Let the distance between the two planes = h m
Given that, AD = 3125 m and ∠ACB = 60° and ∠ACD = 30°

Question 17.
A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60° ; when he retires 20 m from the bank, he finds the angle to be 30°. Find the height of the tree and the breadth of the river. .
Solution:
Let TR be the tree and PR be the width of the river.

Question 18.
The shadow of a vertical tower on a level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. Find the height of the tower, correct to two decimal places. (2006)
Solution:
In the figure, AB is the tower,
BD and BC are the shadow of the tower in two situations.
Let BD = x m and AB = h m
In ∆ABD,

Question 19.
From the top of a hill, the angles of depression of two consecutive kilometer stones, due east are found to be 30° and 45° respectively. Find the distance of two stones from the foot of the hill.
Solution:
Let A and B be the position of two consecutive kilometre stones.
Then AB = 1 km = 1000m
Let the dIstance BC = x m
∴ Distance AC = (1000 + x) m

Question 20.
A man observes the angles of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m the angle of elevation changes to 60°. Find the height of the building correct to the nearest me he.
Solution:
Given that
AB is a building CD = 60 m

Question 21.
At a point on level ground, the angle,of elevation of a vertical lower is found to be such that its tangent is \(\\ \frac { 5 }{ 12 } \). On walking 192 m towards the tower,the tangent of the angle is found to be \(\\ \frac { 3 }{ 4 } \). Find the height of the tower. (1990)
Solution:
Let TR be the tower and P is the point on the
ground such that tan θ = \(\\ \frac { 5 }{ 12 } \)

Question 22.
In the figure, not drawn to scale, TF is a tower. The elevation of T from A is x° where tan x = \(\\ \frac { 2 }{ 5 } \) and AF = 200 m. The elevation of T from B, where AB = 80 m, is y°. Calculate :
(i) The height of the tower TF.
(ii) The angle y, correct to the nearest degree. (1997)

Solution:
Let height of the tower TF = x
tan x = \(\\ \frac { 2 }{ 5 } \), AF = 200 m, AB = 80 m
(i) In right ∆ATF,

Question 23.
From the top of a church spire 96 m high, the angles of depression of two vehicles on a road, at the same level as the base of the spire and on the same side of it are x° and y°, where tan x° = \(\\ \frac { 1 }{ 4 } \) and tan y° = \(\\ \frac { 1 }{ 7 } \). Calculate the distance between the vehicles. (1994)
Solution:
Height of the church CH.
Let A and B are two vehicles which make the angle of depression
from C are x° and y° respectively.

Question 24.
In the adjoining figure, not drawn to the scale, AB is a tower and two objects C and D are located on the ground, on the same side of AB. When observed from the top A of the tower, their angles of depression are 45° and 60°. Find the distance between the two objects. If the height of the tower is 300 m. Give your answer to the nearest metre. (1998)

Solution:
Let CB = x and
DB = y
AB = 300 m

Question 25.
The horizontal distance between two towers is 140 m. The angle of elevation of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 60 m, find the height of the first tower.
Solution:
Let the height of first tower TR = x
height of second tower PQ = 60 m
Distance between the two towers QR = 140 m

Question 26.
As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships on the same side of the ,lighthouse in horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest metre.
Solution:
Let AB be the lighthouse and C and D be the two ships.

Question 27.
The angle of elevation of a pillar from a point A on the ground is 45° and from a point B diametrically opposite to A and on the other side of the pillar is 60°. Find the height of the pillar, given that the distance between A and B is 15 m.
Solution:
Let CD be the pillar and let CD = x
Angles of elevation of points A and B are 45° and 60° respectively.

Question 28.
From two points A and B on the same side of a building, the angles of elevation of the top of the building are 30° and 60° respectively. If the height of the building is 10 m, find the distance between A and B correct to two decimal places
Solution:
In ∆DBC, tan 60° = \(\\ \frac { 10 }{ BC } \)
⇒ √3= \(\\ \frac { 10 }{ BC } \)
⇒ BC = \(\frac { 10 }{ \sqrt { 3 } } \)
∆DBC ,tan 30° = \(\\ \frac { 10 }{ BC+AB } \)

Question 29.
(i) The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the; two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number. (2017)
(ii) An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number. (2014)
Solution:
(i) Let AD be the height of the lighthouse CD = 60 m
Let AD = x m, BD = y m

Question 30.
From a tower 126 m high, the angles of depression of two rocks which are in a horizontal line through the base of the tower are 16° and 12°20′ Find the distance between the rocks if they are on
(i) the same side of the tower
(ii) the opposite sides of the tower.
Solution:
Let CD be the tower and CD = 126 m
Let A and B be the two rocks on the same line
and angles of depression are 16° and 12°20′ respectively,

Question 31.
A man 1.8 m high stands at a distance of 3.6 m from a lamp post and casts a shadow of 5.4 m on the ground. Find the height of the lamp post.
Solution:
AB is the lamp post CD is the height of man.
BD is the distance of man from the foot of the lamp
and FD is the shadow of man.
CE || DB.

Question 32.
The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45° respectively. Find the height of tire multi-storeyed building and the distance between the two buildings, correct to two decimal places.
Solution:
Let AB be the CD be the building
The angles of depression in from A, to C
and D are 30° and 45° respectively
∠ACE = 30° and ∠ADB = 45°
CD = 8 m

Question 33.
A pole of height 5 m is fixed on the top of a tower. The angle of elevation of the top of the pole as observed from a point A on the ground is 60° and the angle of depression of the point A from the top of the tower is 45°. Find the height of the tower. (Take √3 = 1.732).
Solution:
Let QR be the tower and PQ be the pole on it
Angle of elevation from P to a point A is ∠PAR = 60°
and angle of depression from Q to A = 45°
∠QAR = 45° (alternate angle)
PQ = 5 m,

Question 34.
A vertical pole and a vertical tower are on the same level ground. From the top of the pole the angle of elevation of the top of the tower is 60° and the angle of depression of the foot of the tower is 30°. Find the height of the tower if the height of the pole is 20 m.
Solution:
Let TR is tower and
PL is the pole on the same level, ground PL = 20m
From P, draw PQ || LR
then ∠ TPQ = 60° and ∠ QPR = 30°

Question 35.
From the top of a building 20 m high, the angle of elevation of the top of a monumenti is 45° and the angle of depression of its foot is 15°. Find the height of the monument.
Solution:
Let AB be the building and AB = 20 m and
let CD be the monument and let CD = x
The distance between the building and the monument be y,

Question 36.
The angle of elevation of the top of an unfinished tower at a point distant 120 m from its base is 45°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60°?
Solution:
Let AB be the unfinished tower and AB = 120 m
and angle of elevation = 45°
Let x be higher raised so that
the angle of elevation becomes 60°

Question 37.
In the adjoining figure, the shadow of a vertical tower on the level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. Find the height of the tower and give your answer, correct to \(\\ \frac { 1 }{ 10 } \) of a metre.

[Remark. Altitude of the sun means angle of elevation of the sun.]
Solution:
Let TR be the tower and TR = h ;
Let BR = x,
AB = 10 m
Angles of elevation from the top of the tower
at A and B are 30° and 45° respectively.

Question 38.
An aircraft is flying at a constant height with a speed of 360 km/h. From a point on the ground, the angle of elevation of the aircraft at an instant was observed to be 45°. After 20 seconds, the angle of elevation was observed to be 30°. Determine the height at which the aircraft is flying (use √3 = 1.732)
Solution:
Speed of aircraft = 360 km/h
Distance covered in 20 seconds = \(\\ \frac { 360X20 }{ 60X60 } \) = 2 km
E is the fixed point on the ground
and CD is the position of AB in height of aircraft

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

Question 1.
Using trigonometrical tables, find the values of :
(i) sin 48° 52′
(ii) cos 37° 34′
(iii) tan 18° 21′.
Solution:
Using tables, we find that
(i) sin 48° 52′ = .7524 + .0008 = .7532
(ii) cos 37° 34′ = .7934 – .0007 = .7927
(iii) tan 18° 21′ = .3307 + .0010 = .3317.

Question 2.
Use tables to find the acute angle θ, given that
(i) sin θ = 0.5766
(ii) cos θ = 0.2495
(iii) tan θ = 2.4523.
Solution:
Using table, we find that
(i) sin θ = 0.5766 = 0.5764 + 0.0002
= sin (35° 12’+ 1′)
= sin 35° 13′
θ = 35° 13′
(ii) cos θ = 0.2495 = 0.2487 + 0.0008
= cos (75° 36′ – 3′)
= cos 75° 33′
θ = 75° 33′
(iii) tan θ = 2.4523 = 2.4504 + 0.0019
= tan (67° 48′ + 1′)
= tan 67° 49′

Question 3.
If θ is acute and cos θ = 0.53, find the value of tan θ.
Solution:
From the table, we find that
cos θ = 0.53 = .5299 + .0001 = cos 58°
θ = 58°
and tan 58° = 1.6003

Question 4.
Find the value of: sin 22° 11′ + cos 57° 20′ – 2 tan 9° 9′.
Solution:
Using the tables, we find that
sin 22° 11′ = 0.3762 + 0.0014 = 0.3776
cos 57° 20′ = 0.5402 – 0.0005 = 0.5397
tan 9° 9′ = 0.1602 + 0.0009 = 0.1611
∴ sin 22° 11′ + cos 57° 20′ – 2 tan 9° 9′
= 0.3376 + 0.5397 – 0.1611 × 2
= 0.3776 + 0.5397 – 0.3222
= 0.9173 – 0.3222
= .5951.

Question 5.
If θ is acute and sin θ = 0.7547, find the value of: (i) θ (ii) cos θ (iii) 2 cos θ – 3 tan θ.
Solution:
Using the tables, we find that
(i) sin θ = 0.7547 = sin 49°
θ = 49°.
(ii) cos θ = cos 49° = 0.6561.
(iii) tan θ = tan 49° = 1.1504
2 cos θ – 3 tan θ
= 2 × θ.6561 – 3 × 1.1504
= 1.3122 – 3.4512
= – 2.1390

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

Question 1.
Find the value of the following:
(i) sin 35° 22′
(ii) sin 71° 31′
(iii) sin 65° 20′
(iv) sin 23° 56′.
Solution:
(i) sin 35° 22′
Using the table of natural sines,
we see 35° in the horizontal line and for 18′,
in the vertical column, the value is 0.5779.
Now read 22′ – 18′ = 4′ in the difference column, the value is 10.
Adding 10 in 0.5779 + 10 = 0.5789,
we find sin 35° 22′ = 0.5789.
(ii) sin 71° 31′
Using the table of natural sines, we see 71° in the horizontal line
and for 30′ in the vertical column, the value is 0.9483 and for 31′ – 30′ = 1′,
we see in the mean difference column, the value is 1.
∴ sin 71° 31′ = 0.9483 + 1 = 0.9484.
(iii) sin 65° 20′
Using the table of natural sines, we see 65° in the horizontal line
and for 18′ in the vertical column, the value is .9085 and for 20′ – 18′ = 2′,
we see in the mean difference column. We find 2.
∴ sin 65° 20′ = 0.9085 + 2 = 0.9087 Ans.
(iv) sin 23° 56′
Using the table of natural sines, we see 23° in the horizontal line
and for 54′, we see in vertical column, the value is 0.4051
and for 56′ – 54′ = 2′ in the mean difference. It is 5.
∴ sin 23° 56′ = 0.4051 + 5 = 0.4056

Question 2.
Find the value of the following:
(i) cos 62° 27′
(ii) cos 3° 11′
(iii) cos 86° 40′
(iv) cos 45° 58′.
Solution:
(i) cos 62° 27′
From the table of natural cosines,
we see 62° in the horizontal line and 24′ in the vertical column, the value is .4633
and 27′ – 24′ = 3′ in the mean difference. Its value is 8.
∴ cos 62° 27′ = 0.4633 – 8 = 0.4625 Ans.
(ii) cos 3° 11′
From the table of natural cosines, we see 3° in the horizontal line
and 6′ in the vertical column, its value is 0.9985
and 11′ – 6′ = 5′ in the mean difference, its value is 1.
∴ cos 3° 11′ = 0.9985 – 1 = 0.9984 Ans.
(iii) cos 86° 40′
From the table of natural cosines, we see 86° in the horizontal line
and 36′ in the vertical column, its value is 0.0593
and for 40′ – 36′ = 4′ in the mean difference, it is 12.
cos 86° 40’= 0.0593 – 12 = 0 0581 Ans.
(iv) cos 45° 58′
From the table of natural cosines, we see 45° in the horizontal column
and 54′ in the vertical column, its value is 0.6959
and for 58′ – 54′ = 4′, in the mean difference, it is 8.’
cos 45° 58′ = 0.6959 – 8 = 0.6951

Question 3.
Find the value of the following :
(i) tan 15° 2′
(ii) tan 53° 14′
(iii) tan 82° 18′
(iv) tan 6° 9′.
Solution:
(i) tan 15° 2′
From the table of natural tangents, we see 15° in the horizontal line,
its value is 0.2679 and for 2′, in the mean difference, it is 6.
tan 15° 2′ = 0.2679 + 6 = 0.2685.
(ii) tan 53° 14′
From the table of natural tangents, we see 53° in the horizontal line
and 12′ in the vertical column, its value is 1.3367
and 14′ – 12′ = 2′ in the mean difference, it is 16.
∴ tan 53° 14′ = 1.3367 + 16 = 1 .3383 Ans.
(iii) tan 82° 18′
From the table of natural tangents, we see 82° in the horizontal line
and 18′ in the vertical column, its value is 7.3962.
∴ tan 82° 18’= 7.3962.
(iv) tan 6° 9′
From the table of natural tangents, we see 6° in the horizontal line
and 6′ in the vertical column, its value is .1069
and 9′ – 6′ = 3′, in the mean difference, it is 9.
tan 6°9′ = .1069 + 9 = .1078.

Question 4.
Use tables to find the acute angle θ, given that:
(i) sin θ = – 5789
(ii) sin θ = – 9484
(iii) sin θ = – 2357
(iv) sin θ = – 6371.
Solution:
(i) sin θ = – 5789
From the table of natural sines,
we look for the value (≤ 5789), which must be very close to it,
we find the value .5779 in the column 35° 18′ and in mean difference,
we see .5789 – .5779 = .0010 in the column of 4′.
θ = 35° 18’+ 4’= 35° 22′ Ans.
(ii) sin θ = . 9484
From the table of natural sines,
we look for the value (≤ 9484) which must be very close to it,
we find the value .9483 in the column 71° 30′
and in the mean differences,
we see .9484 – 9483 = 0001, in the column of 1′.
θ = 71° 30′ + 1′ = 71° 31′ Ans.
(iii) sin θ = – 2357
From the table of natural sines,
we look for the value (≤ 2357) which must be very close to it,
we find the value .2351 in the column 13° 36′ and in the mean difference,
we see .2357 – 2351 = .0006, in the column of 2′.
θ = 13° 36′ +2’= 13° 38′ Ans.
(iv) sin θ = .6371
From the table of natural sines,
we look for the value (≤ 6371) which must be very close to it,
we find the value .6361 in the column 39° 30′ and in the mean difference,
we see .6371 – .6361 = .0010 in the column of 4′.
θ = 39° 30′ + 4′ = 39° 34′

Question 5.
Use the tables to find the acute angle θ, given that:
(i) cos θ = .4625
(ii) cos θ = .9906
(iii) cos θ = .6951
(iv) cos θ = .3412.
Solution:
(i) cos θ = .4625
From the table of natural cosines,
we look for the value (≤ .4625) which must be very close to it,
we find the value .4617 in the column of 62° 30′ and in the mean difference,
we see .4625 – .4617 = .0008 which is in column of 3′.
θ = 62° 30′ – 3’= 62° 27′.
(ii) cos θ = .9906
From the table of cosines,
we look for the value (≤ .9906) which must be very close to it,
we find the value of .9905 in the column of 7° 54′ and in mean difference,
we see .9906 – 9905 = .0001 which is in column of 3′.
θ = 70 54′ – 3’= 7° 51′
(iii) cos θ = .6951
From the tables of cosines,
we look for the value (≤ 6951) which must be very close to it,
we find the value .6947 in the column of 46° and in mean difference,
.6951 – .6947 = 0.0004 which in the column of 2′.
θ = 46° – 2′ = 45° 58′ Ans.
(iv) cos θ = .3412
From the table of cosines,
we look for the value of (≤ .3412) which must be very close to it,
we find the value .3404 in the column of 70° 6′ and in the mean difference,
.3412 – 3404 = .0008 which is in the column of 3′.
θ = 70° 6′ – 3′ = 70° 3′

Question 6.
Use tables to find the acute angle θ, given that:
(i) tan θ = .2685
(ii) tan θ = 1.7451
(iii) tan θ = 3.1749
(iv) tan θ = .9347
Solution:
(i) tan θ = .2685
From the table of natural tangent,
we look for the value of (≤ .2685) which must be very close to it,
we find the value .2679 in the column of 15° and in the mean difference,
.2685 – .2679 which is in the column of 2′.
θ = 15° +2′ = 15° 2′ Ans.
(ii) tan θ = 1.7451
From the tables of natural tangents,
we look for the value of (≤ 1.7451) which must be very close to it,
we find the value 1.7391 in the column of 60°’ 6′
and in the mean difference 1.7451 + 1.7391 = 0.0060 which is in the column of 5′.
θ = 60° 6’+ 5’= 60° 11’Ans.
(iii) tan θ = 3.1749
From the tables of natural tangents,
we look for the value of (≤ 3.1749) which must be very close to it,
we find the value 3.1716 in the column of 72° 30′
and in the mean difference 3.1749 – 3.1716 = 0.0033 which is in the column of 1′.
θ = 720 30′ + 1′ = 72° 31′ Ans.
(iv) tan θ = .9347
From the tables of natural tangents,
we look for the value of (≤ .9347 which must be very close to it,
we find the value .9325 in the column of 43°
and in the mean difference .9347 – .9325 = 0.0022 which is in the column of 4′.
θ = 43° + 4′ = 43° 4′

Question 7.
Using trigonometric table, find the measure of the angle A when sin A = 0.1822.
Solution:
sin A = 0.1822
From the tables of natural sines,
we look for the value (≤ .1822) which must be very close to it,
we find the value .1822 in column 10° 30′.
A = 10° 30′

Question 8.
Using tables, find the value of 2 sin θ – cos θ when (i) θ = 35° (ii) tan θ = .2679.
Solution:
(i) θ = 35°
2 sin θ – cos θ = 2 sin 35° – cos 35°
= 2 x .5736 – .8192
(From the tables)
= 1.1472 – .8192 = 0.3280.
(ii) tan θ = .2679
From the tables of natural tangents,
we look for the value of ≤ .2679,
we find the value of the column 15°.
θ = 15°
Now, 2 sin θ – cos θ = 2 sin 15° – cos 15°
= 2 (.2588) – .9659 = 5136 – .9659
= -0.4483

Question 9.
If sin x° = 0.67, find the value of
(i) cos x°
(ii) cos x° + tan x°.
Solution:
sin x° = 0.67
From the table of natural sines,
we look for the value of (≤ 0.67) which must be very close to it,
we find the value .6691 in the column 42° and in the mean difference,
the value of 0.6700 – 0.6691 = 0.0009 which is in the column 4′.
θ = 42° + 4′ = 42° 4′
Now
(i) cos x° = cos 42° 4′ = .7431 – .0008
= 0.7423 Ans.
(ii) cos x° + tan x° = cos 42° 4′ + tan 42° 4′
= 0.7423 + .9025
= 1.6448

Question 10.
If θ is acute and cos θ = .7258, find the value of (i) θ (ii) 2 tan θ – sin θ.
Solution:
cos θ = .7258
From the table of cosines,
we look for the value of (≤ .7258) which must be very close to it,
we find the value .7254 in the column of 43° 30′
and in the mean differences the value of .7258 – .7254 = 0.0004
which in the column of 2′.
(i) θ = 43° 30′ – 2’= 43° 28′.
(ii) 2 tan θ – sin θ
= 2 tan43°28′ – sin43°28′
= 2 (.9479) – .6879
= 1.8958 – .6879
= 1.2079

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

Question 1.
(i) If θ is an acute angle and cosec θ = √5 find the value of cot θ – cos θ.
(ii) If θ is an acute angle and tan θ = \(\\ \frac { 8 }{ 15 } \), find the value of sec θ + cosec θ.
Solution:
(i) θ is an acute angle.
cosec θ = √5

Question 2.
Evaluate the following:
(i) \(2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right) \) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°
(ii) \(\frac { { sec }29^{ O } }{ { cosec }61^{ O } } \) + 2 cot 8° cot 17° cot 45° cot 73°0 cot 82° – 3(sin2 38° + sin2 52°)
(iii) \(\frac { { sin }^{ 2 }{ 22 }^{ O }+{ sin }^{ 2 }{ 68 }^{ O } }{ { cos }^{ 2 }{ 22 }^{ O }+{ cos }^{ 2 }{ 68 }^{ O } } \) + sin2 63° + cos 63° sin 27°
Solution:
(i) \(2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right) \) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

Question 3.
If \(\\ \frac { 4 }{ 3 } \) (sec2 59° – cot2 31°) – \(\\ \frac { 2 }{ 2 } \) sin 90° + 3tan2 56° tan2 34° = \(\\ \frac { x }{ 2 } \), then find the value of x.
Solution:
Given
\(\\ \frac { 4 }{ 3 } \) (sec2 59° – cot2 31°) – \(\\ \frac { 2 }{ 2 } \) sin 90° + 3tan2 56° tan2 34° = \(\\ \frac { x }{ 2 } \)

Prove the following (4 to 11) identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

Question 4.
(i) \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA \)
(ii) \(\frac { cosA }{ cosecA+1 } +\frac { cosA }{ cosecA-1 } =2tanA \)
Solution:
(i) \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA \)
L.H.S = \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } \)

Question 5.
(i) \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta \)
(ii) (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1.
Solution:
(i) \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta \)
L.H.S = \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } \)

Question 6.
(i) sin² θ + cos⁴ θ = cos² θ + sin⁴ θ
(ii) \(\frac { cot\theta }{ cosec\theta +1 } +\frac { cosec\theta +1 }{ cot\theta } =2sec\theta \)
Solution:
L.H.S. = sin² θ + cos⁴ θ
= (1 – cos² θ + cos⁴ θ
= 1 – cos² θ + cos⁴ θ
= 1 – cos² θ (1 – cos² θ)

Question 7.
(i) sec⁴ A (1 – sin⁴ A) – 2 tan² A = 1
(ii) \(\frac { 1 }{ sinA+cosA+1 } +\frac { 1 }{ sinA+cosA-1 } =secA+cosecA\)
Solution:
(i) sec⁴ A (1 – sin⁴ A) – 2 tan² A = 1
L.H.S = sec⁴ A (1 – sin⁴ A) – 2 tan² A

Question 8.
(i) \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1\)
(ii) (sec A – tan A)² (1 + sin A) = 1 – sin A.
Solution:
(i) \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1\)
L.H.S = \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta\)

Question 9.
(i) \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA \)
(ii) (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A
(iii) \(\frac { { tan }^{ 2 }\theta }{ { tan }^{ 2 }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { sec }^{ 2 }\theta -{ cosec }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta } \)
Solution:
(i) \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA \)
L.H.S = \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } \)

Question 10.
\(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 } \)
Solution:
\(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 } \)
L.H.S = \(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } \)

Question 11.
2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1 = θ
Solution:
2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1 = θ
L.H.S = 2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1

Question 12.
If cot θ + cos θ = m, cot θ – cos θ = n, then prove that (m² – n²)² = 16 run.
Solution:
cot θ + cos θ = m…..(i)
cot θ – cos θ = n……(ii)
Adding (i)&(ii) we get

Question 13.
If sec θ + tan θ = p, prove that sin θ = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)
Solution:
sec θ + tan θ = p,
prove that sin θ = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)
\(\frac { 1 }{ cos\theta } +\frac { sin\theta }{ cos\theta } =p\)

Question 14.
If tan A = n tan B and sin A = m sin B, prove that cos² A = \(\frac { { m }^{ 2 }-1 }{ { n }^{ 2 }-1 } \)
Solution:
m = \(\\ \frac { sinA }{ sinB } \)
n = \(\\ \frac { tanA }{ tanB } \)

Question 15.
If sec A = \(x+ \frac { 1 }{ 4x } \), then prove that sec A + tan A = 2x or \(\\ \frac { 1 }{ 2x } \)
Solution:
sec A = \(x+ \frac { 1 }{ 4x } \)
To prove that sec A + tan A = 2x or \(\\ \frac { 1 }{ 2x } \)

Question 16.
When 0° < θ < 90°, solve the following equations:
(i) 2 cos² θ + sin θ – 2 = 0
(ii) 3 cos θ = 2 sin² θ
(iii) sec² θ – 2 tan θ = 0
(iv) tan² θ = 3 (sec θ – 1).
Solution:
0° < θ < 90°
(i) 2 cos² θ + sin θ – 2 = 0

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

Choose the correct answer from the given four options (1 to 12) :

Question 1.
\({ cot }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta } \) is equal to
(a) 1
(b) -1
(c) sin² θ
(d) sec² θ
Solution:
\({ cot }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta } \)
= \(\frac { { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta } -\frac { 1 }{ { sin }^{ 2 }\theta } \)

Question 2.
(sec² θ – 1) (1 – cosec² θ) is equal to
(a) – 1
(b) 1
(c) 0
(d) 2
Solution:
(sec² θ – 1) (1 – cosec² θ)
= \(\left( \frac { 1 }{ { cos }^{ 2 }\theta } -1 \right) \left( 1-\frac { 1 }{ { sin }^{ 2 }\theta } \right) \)

Question 3.
\(\frac { { tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \) is equal to
(a) 2 sin² θ
(b) 2 cos² θ
(c) sin² θ
(d) cos² θ
Solution:
\(\frac { { tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \)

= \(=\frac { { sin }^{ 2 }\theta }{ 1 } ={ sin }^{ 2 }\theta \) (c)

Question 4.
(cos θ + sin θ)² + (cos θ – sin θ)² is equal to
(a) – 2
(b) 0
(c) 1
(d) 2
Solution:
(cos θ + sin θ)² + (cos θ – sin θ)²
= cos² θ + sin² θ + 2 sin θ cos θ + cos² θ + sin² θ – 2 sin θ cos θ
= 2(sin² θ + cos² θ)
= 2 × 1 = 2 (d)
(∵ sin² θ + cos² θ = 1)

Question 5.
(sec A + tan A) (1 – sin A) is equal to
(a) sec A
(b) sin A
(c) cosec A
(d) cos A
Solution:
(sec A + tan A) (1 – sin A)

Question 6.
\(\frac { { 1+tan }^{ 2 }A }{ { 1+cot }^{ 2 }A } \) is equal to
(a) sec² A
(b) – 1
(c) cot² A
(d) tan² A
Solution:
\(\frac { { 1+tan }^{ 2 }A }{ { 1+cot }^{ 2 }A } \)

Question 7.
If sec θ – tan θ = k, then the value of sec θ + tan θ is
(a) \(1-\frac { 1 }{ k } \)
(b) 1 – k
(c) 1 + k
(d) \(\\ \frac { 1 }{ k } \)
Solution:
sec θ – tan θ = k
\(\frac { 1 }{ cos\theta } -\frac { sin\theta }{ cos\theta } =k\)

Question 8.
Which of the following is true for all values of θ (0° < θ < 90°):
(a) cos² θ – sin² θ = 1
(b) cosec² θ – sec² θ = 1
(c) sec² θ – tan² θ = 1
(d) cot² θ – tan² θ = 1
Solution:
∴ sec² θ – tan² θ = 1 is true for all values of θ as it is an identity.
(0° < θ < 90°) (c)

Question 9.
If θ is an acute angle of a right triangle, then the value of sin θ cos (90° – θ) + cos θ sin (90° – θ) is
(a) 0
(b) 2 sin θ cos θ
(c) 1
(d) 2 sin² θ
Solution:
sin θ cos (90° – θ) + cos θ sin (90° – θ)
= sin θ sin θ + cos θ cos θ
{ ∵ sin(90° – θ) = cosθ, cos (90° – θ) = sin θ }
= sin² θ + cos² θ = 1 (c)

Question 10.
The value of cos 65° sin 25° + sin 65° cos 25° is
(a) 0
(b) 1
(b) 2
(d) 4
Solution:
cos 65° sin 25° + sin 65° cos 25°
= cos (90° – 25°) sin 25° + sin (90° – 25°) cos 25°
= sin 25° . sin 25° + cos 25° . cos 25°
= sin² 25° + cos² 25°
( ∵ sin² θ + cos² θ = 1)
= 1 (b)

Question 11.
The value of 3 tan² 26° – 3 cosec² 64° is
(a) 0
(b) 3
(c) – 3
(d) – 1
Solution:
3 tan² 26° – 3 cosec² 64°
= 3 tan² 26° – 3 cosec (90° – 26°)
= 3 tan² 26° – 3 sec² 26°

Question 12.
The value of \(\frac { sin({ 90 }^{ O }-\theta )sin\theta }{ tan\theta } -1 \) is
(a) – cot θ
(b) – sin² θ
(c) – cos² θ
(d) – cosec² θ
Solution:
\(\frac { sin({ 90 }^{ O }-\theta )sin\theta }{ tan\theta } -1 \)

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS, drop a comment below and we will get back to you at the earliest.