# ML Aggarwal Solutions Class 10 Maths Chapter 19 Trigonometric Tables

1. Find out  the value of the following:
(i) sin 35o 22′
Solution:-
To find the value of sin 35o 22’,
When We read the table of natural sines in the horizontal line which begins with 35o
In the vertical column headed by 22’ i.e. 22’ – 18’ = 4’
in the difference column,
then the value of 4’ in mean difference column is 10.
Then, value that we find in vertical column is 0.5779
Therefore adding the value of 18’ and 4’ = 0.5779 + 10
= 0.5789
Therefore, the value of sin 35o 22’ is obtained as under,
sin 35o 22’ = 0.5779                 … [from table]
Mean difference for 4’ = 10        … [to be added]
Therefore, sin 35o 22’ = 0.5789

(ii) sin 71o 31′
Solution:-
To find out the value of sin 71o 31’,
When We read the table of natural sines in the horizontal line which begins with 35o
In the vertical column headed by 31’ i.e. 31’ – 30’ = 1’
in the difference column,
the value of 1’ in mean difference column is 1.
Therefore , value that we find in vertical column is 0.9483
Now adding the value of 30’ and 1’ = 0.9483 + 1
= 0.9484

Therefore, the value of sin 71o 31’ is obtained as under,
sin 71o 31’ = 0.9483              … [from table]
Mean difference for 1’ = 1      … [to be added]
Then, sin 71o 31’ = 0.9484

(iii) sin 65o 20′
Solution:-
To find out the value of sin 65o 20’,
When We read the table of natural sines in the horizontal line which begins with 35
In the vertical column headed by 20’ i.e. 20’ – 18’ = 2’
in the difference column,
the value of 2’ in mean difference column is 2.
Therefore  , value that we find in vertical column is 0.9085
Now adding the value of 18’ and 2’ = 0.9085 + 2
= 0.9087
Therefore, the value of sin 65o 20’ is obtained as under,
sin 65o 20’ = 0.0985            … [from table]
Mean difference for 2’ = 2           … [to be added]
Hence, sin 65o 20’ = 0.9087

(iv) sin 23o 56′
Solution:-
To find out the value of sin 23o 56’,
When  We read the table of natural sines in the horizontal line which begins with 23o
In the vertical column headed by 56’ i.e. 56’ – 54’ = 2’ in the difference column,
the value of 2’ in mean difference column is 5.
Then,
value that we find in vertical column is 0.4051
Therefore adding the value of 54’ and 4’ = 0.4051 + 5
= 0.4056
Therefore, the value of sin 23o 56’ is obtained as under,
sin 23o 56’ = 0.4051          …   [from table]
Mean difference for 2’ = 5            … [to be added]
Hence , sin 23o 56’ = 0.4056

2. Find out the value of the following:
(i) cos 62o 27′
Solution:-
As We know that as θ increase, the value of cos θ decreases, therefore, the numbers in the mean difference columns are to be subtracted.
To find the value of cos 62o 27’,
When  We read the table of natural sines in the horizontal line which begins with 62o
In the vertical column headed by 27’ i.e. 27’ – 24’ = 3’ in the difference column,
the value of 3’ in mean difference column is 8.
Then, value that we find in vertical column is 0.4633
Now adding the value of 24’ and 3’ = 0.4633 - 8
= 0.4625
Hence , cos 62o 27’ is 0.4625.

(ii) cos 3° 11′
Solution:-
As We know that as θ increase,
the value of cos θ decreases, therefore, the numbers in the mean difference columns are to be subtracted.
To find out  the value of cos 3o 11′,
When We read the table of natural sines in the horizontal line which begins with 3o
In the vertical column headed by 27’ i.e. 11’ – 6’ = 5’
in the difference column, the value of 5’ in mean difference column is 1.
Then, value that we find in vertical column is 0.9985
Now adding the value of 6’ and 5’ = 0.9985 - 1
= 0.9984
Hence , cos 3o 11′ is 0.9984.

(iii) cos 86o 40′
Solution:-
As  We know that as θ increase, the value of cos θ decreases, therefore, the numbers in the mean difference columns are to be subtracted.
To find the value of cos 86o 40′,
When We read the table of natural sines in the horizontal line which begins with 86o
In the vertical column headed by 40’ i.e. 40’ – 36’ = 4’
in the difference column,
the value of 4’ in mean difference column is 12.
Then, value that we find in vertical column is 0.0593
Therefore  adding the value of 36’ and 4’ = 0.0593 - 12
= 0.0581
Hence , cos 86° 40′ is 0.0581.

(iv) cos 45o 58′.
Solution:-
As We know that as θ increase, the value of cos θ decreases, therefore, the numbers in the
mean difference columns are to be subtracted.
To find the value of cos 45o 58′,
When We read the table of natural sines in the horizontal line which begins with 45o
In the vertical column headed by 58’ i.e. 58’ – 54’ = 4’
in the difference column,
the value of 4’ in mean difference column is 8.
Therefore , value that we find in vertical column is 0.6959
Now adding the value of 54’ and 4’ = 0.6959 - 8
= 0.6951
Hence , cos 45° 58′ is 0.6951.

3. Find out the value of the following :
(i) tan 15o 2′
Solution:-
To find out the value of tan 15o 2′,
When We read the table of natural sines in the horizontal line which begins with 15o
In the vertical column headed by 2’, the value of 2’ in mean difference column is 6.
Therefore , value that we find in vertical column is 0.2679
Now adding the values = 0.2685 + 6
= 0.2685
Hence, tan 15o 2’ is 0.2685.

(ii) tan 53o 14′
Solution:-
To find out  the value of tan 53o 14′,
When We read the table of natural sines in the horizontal line which begins with 53o In the vertical column headed by 14’ i.e. 14’ – 12’ = 2’ in the difference column, the value of 2’ in mean difference column is 16.
Therefore , value that we find in vertical column is 1.3367
Now adding the value of 12’ and 2’ = 1.3367 + 16
= 1.3383
Hence, tan 53o 14′ is 1.3383.

(iii) tan 82o 18′
Solution:-
To find out the value of tan 82o 18′,
When We read the table of natural sines in the horizontal line which begins with 82o
Then, value that we find in vertical column is 7.3962
Hence , tan 82o 18’ is 7.3962

(iv) tan 6o 9′.
Solution:-
To find out the value of tan 6o 9′,
When We read the table of natural sines in the horizontal line which begins with 6o
In the vertical column headed by 9’ i.e. 9’ – 6’ = 3’
in the difference column,
the value of 3’ in mean difference column is 9.

Therefore , value that we find in vertical column is 0.1069
Now adding the value of 6’ and 3’ = 0.1069 + 9
= 0.1078
Hence , tan 6o 9′ is 0.1078.

4. Use tables to find out  the acute angle θ, given that:
(i) sin θ = .5789
Solution:-
As In the table of natural sines, look for a value (≤ .5789) which is sufficiently close to .5789. We find  out the value .5779 occurs in the horizontal line beginning with 35o and in the column headed by 18’ and in the mean difference,we see .
5789 - .5779 = .0010
in the column of 4’.
So we get,
θ = 35o 18’ + 4’
= 35o 22’

(ii) sin θ = .9484
Solution:-
AsIn the table of natural sines, look for a value (≤ .9484) which is sufficiently close to .9484.
We find out the value .9483 occurs in the horizontal line beginning with 71o and in the column headed by 30’ and in the mean difference, we see .9484 - .9483 = .0001 in the  column of 1’.
Then  we get,
θ = 71o 30’ + 1’
= 71o 31’.

(iii) sin θ = .2357
Solution:-
As In the table of natural sines, look for a value (≤ .2357) which is sufficiently close to .2357.
We find  out  the value .2351 occurs in the horizontal line beginning with 13o and in the column headed by 36’ and in the mean difference, we see .2357 - .2351 = .0006 in the column of 2’.
Then we get,
θ = 13o 36’ + 2’ = 13o 38’

(iv) sin θ = .6371.
Solution:-
As In the table of natural sines, look for a value (≤ .6371) which is sufficiently close to .6371.
We find out the value .6361 occurs in the horizontal line beginning with 39o and in the column headed by 30’ and in the mean difference, we see .6371 - .6361 = .0010 in the column of 4’.
Then we get, θ = 39o 30’ + 4’ = 39o 34’.

5. Use the tables to find out  the acute angle θ, given that:
(i) cos θ = .4625
Solution:-
As In the table of cosines, look for a value (≤ .4625) which is sufficiently close to .4625.
We find  out  the value .4617 occurs in the horizontal line beginning with 62o and in the column headed by 30’ and in the mean difference, we see .4625 - .4617 = .0008 in the column of 3’.
Then,  we get, θ = 62o 30’ - 3’
= 62o 27’.

(ii) cos θ = .9906
Solution:-
As In the table of cosines, look for a value (≤ .9906) which is sufficiently close to .9906. We find out the value .9905 occurs in the horizontal line beginning with 7o and in the column headed by 54’ and in the mean difference,
we see .9906 - .9905 = .0001
in the column of 3’.
Then we get,
θ = 70o 54’ - 3’ = 70o 51’.

(iii) cos θ = .6951
Solution:-
As In the table of cosines, look for a value (≤ .6951) which is sufficiently close to .6951.
We find out  the value .6947 occurs in the horizontal line beginning with 46o and in the mean difference, we see .
6951 - .6947 = .0004 in the column of 2’.
Then  we get,
θ = 46o ’ - 2’ = 45o 58’.

(iv) cos θ = .3412.
Solution:-
As In the table of cosines, look for a value (≤ .3412)
which is sufficiently close to .3412.
We find out the value .3404 occurs in the horizontal line beginning with 70o and in the column headed by 6’ and in the mean difference,
we see .3412 - .3404 = .0008 in the column of 3’.
Then  we get, θ = 70o 6’ - 3’ = 70o 3’.

6. Use tables to find  out the acute angle θ, given that:
(i) tan θ = .2685
Solution:-
As In the table of natural tangent, look for a value (≤ .2685) which is sufficiently close to 2685.
We find out the value .2679 occurs in the horizontal line beginning with 15o and in the mean difference,
we see .2685 - .2679 = .0006 in the column of 2’.
Then  we get, θ = 15o + 2’ = 15o 2’.

(ii) tan θ = 1.7451
Solution:-
As In the table of natural tangent, look for a value (≤ 1.7451) which is sufficiently close to 1.7451. We find out  the value 1.7391 occurs in the horizontal line beginning with 60o and in the column headed by 6’ and in the mean difference, we see 1.7451 - 1.7391 = .0060 in the column of 5’.
Then we get,
θ = 60o 6’ + 5’ = 60o 11’.

(iii) tan θ = 3.1749
Solution:-
In the table of natural tangent, look for a value (≤ 3.1749) which is sufficiently close to 3.1749.
We find out  the value 3.1716 occurs in the horizontal line beginning with 72o and in the column headed by 30’ and in the mean difference, we see 3.1749 - 3.1716 = .0033 in the column of 1’. Then  we get,
θ = 72o 30’ + 1’ = 72o 31’.

(iv) tan θ = .9347
Solution:-
As In the table of natural tangent, look for a value (≤ .9347) which is sufficiently close to .9347.
We find out the value .9325 occurs in the horizontal line beginning with 43o and in the mean difference, we see .9347 - .9325 = .0022
in the column of 4’.
Then  we get,
θ = 43o + 4’ = 43o 4’.

7. Using trigonometric table, find  out the measure of the angle A when sin A = 0.1822.
Solution:-
As In the table of natural sines, look for a value (≤ 0.1822) which is sufficiently close to 0.1822.
We find out the value 0.1822 occurs in the horizontal line beginning with 10o and in the column headed by 30’.
Then  we get, A = 10o 30’

8. Using tables, find  out the value of 2 sin θ – cos θ when (i) θ = 35° (ii) tan θ = .2679.
Solution:-
(i) We have to find out the value of 2 sin θ – cos θ
From the question it is given that,
value of θ = 35
So, substitute the value of θ,
= 2 sin 35o – cos 35o
From the table value of sin 35o = .5736 and cos 35o
= .8192
= (2 × .5736) - .8192
= 0.3280

(ii) from the question it is given that, tan θ = .2679
As In the table of natural sines, look for a value (≤ .2679) which is sufficiently close to .2679.
We find  out the value column headed by 15o
Then we get, θ = 15
So, substitute the value of θ,
= 2 sin 15o – cos 15o
From the table value of sin 15o = .2588 and cos 15o = .9659
= (2 × .2588) - .9659
= -0.4483

9. If sin x° = 0.67, find the value of (i) cos x° (ii) cos x° + tan x°.
Solution:-
From the question As we have  given that, sin xo = 0.67.
In the table of natural sines, look for a value (≤ 0.67) which is sufficiently close to 0.67.
We find out the value 0.6691 occurs in the horizontal line beginning with 42o and in the mean difference,
we see
0.6700 - 0.6691 = .0009 in the column of 4’.
Then we get, θ = 42o + 4’ = 42o 4’.
Then,

(i) cos xo = cos 42o 4′
From the table
= .7431 – .0008
= 0.7423

(ii) cos xo + tan x° = cos 42° 4′ + tan 42° 4′
= 0.7423 + .9025
= 1.6448

10. If θ is acute and cos θ = .7258, find out the value of (i) θ (ii) 2 tan θ – sin θ.
Solution:-
From the question, cos θ = .7258
As  In the table of cosines, look for a value (≤ .7258) which is sufficiently close to .7258.
We find out  the value .7254 occurs in the horizontal line beginning with 43o and in the column headed by 30’ and in the mean difference, We see .
7258 - .7254 = .0004
in the column of 2’.
Then we get,
θ = 43° 30’ - 2’ = 43o 28’.
(i) θ = 43° 30′ – 2’
= 43° 28′.

(ii) 2 tan θ – sin θ
By Substitute the value θ, we get
= 2 tan43°28' – sin43°28'
= 2 (.9479) – .6879
= 1.8958 – .6879
= 1.2079
Hence, the value of 2 tan θ – sin θ is 1.2079