# ML Aggarwal Solutions for Class 10 Maths Chapter 19 Trigonometric Tables

ML Aggarwal Solutions for Class 10 Maths Chapter 19 Trigonometric Tables, are provided here. Our solution module employs numerous shortcut tips and practical examples to explain all the exercise questions in a simple and easily understandable language. Our expert tutors at BYJUâ€™S have prepared solutions with simpler steps which can be easily solved by students at ease. These solutions will help you in obtaining knowledge and strong command over the subject. Download pdf of ML Aggarwal Solutions for Class 10 Maths Chapter 19 in their respective links.

Chapter 19 – Trigonometric Tables. The approximate values correct up to four decimal places of sine, cosine and tangent of all angles lying between 0 degree and 90 degree can be found from the tables of natural sines, natural cosines and natural tangents given at the end of the book. These tables are called trigonometrical tables. In this ML Aggarwal Solutions for Class 10 Maths Chapter 19, contains the topics related to Trigonometric Tables and Use of Trigonometric Tables.

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Exercise 19

1. Find the value of the following:
(i) sin 35o 22â€²

Solution:-

To find the value of sin 35o 22â€™,

We read the table of natural sines in the horizontal line which begins with 35o

In the vertical column headed by 22â€™ i.e. 22â€™ â€“ 18â€™ = 4â€™ in the difference column, the value of 4â€™ in mean difference column is 10.

Then, value that we find in vertical column is 0.5779

Now adding the value of 18â€™ and 4â€™ = 0.5779 + 10

= 0.5789

Therefore, the value of sin 35o 22â€™ is obtained as under,

sin 35o 22â€™ = 0.5779 â€¦ [from table]

Mean difference for 4â€™ = 10 â€¦ [to be added]

Then, sin 35o 22â€™ = 0.5789

(ii) sin 71o 31â€²

Solution:-

To find the value of sin 71o 31â€™,

We read the table of natural sines in the horizontal line which begins with 35o

In the vertical column headed by 31â€™ i.e. 31â€™ â€“ 30â€™ = 1â€™ in the difference column, the value of 1â€™ in mean difference column is 1.

Then, value that we find in vertical column is 0.9483

Now adding the value of 30â€™ and 1â€™ = 0.9483 + 1

= 0.9484

Therefore, the value of sin 71o 31â€™ is obtained as under,

sin 71o 31â€™ = 0.9483 â€¦ [from table]

Mean difference for 1â€™ = 1 â€¦ [to be added]

Then, sin 71o 31â€™ = 0.9484

(iii) sin 65o 20â€²

Solution:-

To find the value of sin 65o 20â€™,

We read the table of natural sines in the horizontal line which begins with 35o

In the vertical column headed by 20â€™ i.e. 20â€™ â€“ 18â€™ = 2â€™ in the difference column, the value of 2â€™ in mean difference column is 2.

Then, value that we find in vertical column is 0.9085

Now adding the value of 18â€™ and 2â€™ = 0.9085 + 2

= 0.9087

Therefore, the value of sin 65o 20â€™ is obtained as under,

sin 65o 20â€™ = 0.0985 â€¦ [from table]

Mean difference for 2â€™ = 2 â€¦ [to be added]

Then, sin 65o 20â€™ = 0.9087

(iv) sin 23o 56â€²

Solution:-

To find the value of sin 23o 56â€™,

We read the table of natural sines in the horizontal line which begins with 23o

In the vertical column headed by 56â€™ i.e. 56â€™ â€“ 54â€™ = 2â€™ in the difference column, the value of 2â€™ in mean difference column is 5.

Then, value that we find in vertical column is 0.4051

Now adding the value of 54â€™ and 4â€™ = 0.4051 + 5

= 0.4056

Therefore, the value of sin 23o 56â€™ is obtained as under,

sin 23o 56â€™ = 0.4051 â€¦ [from table]

Mean difference for 2â€™ = 5 â€¦ [to be added]

Then, sin 23o 56â€™ = 0.4056

2. Find the value of the following:
(i) cos 62o 27â€²

Solution:-

We know that as Î¸ increase, the value of cos Î¸ decreases, therefore, the numbers in the mean difference columns are to be subtracted.

To find the value of cos 62o 27â€™,

We read the table of natural sines in the horizontal line which begins with 62o

In the vertical column headed by 27â€™ i.e. 27â€™ â€“ 24â€™ = 3â€™ in the difference column, the value of 3â€™ in mean difference column is 8.

Then, value that we find in vertical column is 0.4633

Now adding the value of 24â€™ and 3â€™ = 0.4633 – 8

= 0.4625

Therefore, cos 62o 27â€™ is 0.4625.

(ii) cos 3o 11â€²

Solution:-

We know that as Î¸ increase, the value of cos Î¸ decreases, therefore, the numbers in the mean difference columns are to be subtracted.

To find the value of cos 3o 11â€²,

We read the table of natural sines in the horizontal line which begins with 3o

In the vertical column headed by 27â€™ i.e. 11â€™ â€“ 6â€™ = 5â€™ in the difference column, the value of 5â€™ in mean difference column is 1.

Then, value that we find in vertical column is 0.9985

Now adding the value of 6â€™ and 5â€™ = 0.9985 – 1

= 0.9984

Therefore, cos 3o 11â€² is 0.9984.

(iii) cos 86o 40â€²

Solution:-

We know that as Î¸ increase, the value of cos Î¸ decreases, therefore, the numbers in the mean difference columns are to be subtracted.

To find the value of cos 86o 40â€²,

We read the table of natural sines in the horizontal line which begins with 86o

In the vertical column headed by 40â€™ i.e. 40â€™ â€“ 36â€™ = 4â€™ in the difference column, the value of 4â€™ in mean difference column is 12.

Then, value that we find in vertical column is 0.0593

Now adding the value of 36â€™ and 4â€™ = 0.0593 – 12

= 0.0581

Therefore, cos 86o 40â€² is 0.0581.

(iv) cos 45o 58â€².

Solution:-

We know that as Î¸ increase, the value of cos Î¸ decreases, therefore, the numbers in the mean difference columns are to be subtracted.

To find the value of cos 45o 58â€²,

We read the table of natural sines in the horizontal line which begins with 45o

In the vertical column headed by 58â€™ i.e. 58â€™ â€“ 54â€™ = 4â€™ in the difference column, the value of 4â€™ in mean difference column is 8.

Then, value that we find in vertical column is 0.6959

Now adding the value of 54â€™ and 4â€™ = 0.6959 – 8

= 0.6951

Therefore, cos 45o 58â€² is 0.6951.

3. Find the value of the following :
(i) tan 15o 2â€²

Solution:-

To find the value of tan 15o 2â€²,

We read the table of natural sines in the horizontal line which begins with 15o

In the vertical column headed by 2â€™, the value of 2â€™ in mean difference column is 6.

Then, value that we find in vertical column is 0.2679

Now adding the values = 0.2685 + 6

= 0.2685

Therefore, tan 15o 2â€™ is 0.2685.

(ii) tan 53o 14â€²

Solution:-

To find the value of tan 53o 14â€²,

We read the table of natural sines in the horizontal line which begins with 53o

In the vertical column headed by 14â€™ i.e. 14â€™ â€“ 12â€™ = 2â€™ in the difference column, the value of 2â€™ in mean difference column is 16.

Then, value that we find in vertical column is 1.3367

Now adding the value of 12â€™ and 2â€™ = 1.3367 + 16

= 1.3383

Therefore, tan 53o 14â€² is 1.3383.

(iii) tan 82o 18â€²

Solution:-

To find the value of tan 82o 18â€²,

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

Therefore, tan 82o 18â€™ is 7.3962

(iv) tan 6o 9â€².

Solution:-

To find the value of tan 6o 9â€²,

We read the table of natural sines in the horizontal line which begins with 6o

In the vertical column headed by 9â€™ i.e. 9â€™ â€“ 6â€™ = 3â€™ in the difference column, the value of 3â€™ in mean difference column is 9.

Then, value that we find in vertical column is 0.1069

Now adding the value of 6â€™ and 3â€™ = 0.1069 + 9

= 0.1078

Therefore, tan 6o 9â€² is 0.1078.

4. Use tables to find the acute angle Î¸, given that:
(i) sin Î¸ = .5789

Solution:-

In the table of natural sines, look for a value (â‰¤ .5789) which is sufficiently close to .5789.

We find the value .5779 occurs in the horizontal line beginning with 35o and in the column headed by 18â€™ and in the mean difference, we see .5789 – .5779 = .0010 in the column of 4â€™.

So we get, Î¸ = 35o 18â€™ + 4â€™ = 35o 22â€™.

(ii) sin Î¸ = .9484

Solution:-

In the table of natural sines, look for a value (â‰¤ .9484) which is sufficiently close to .9484.

We find the value .9483 occurs in the horizontal line beginning with 71o and in the column headed by 30â€™ and in the mean difference, we see .9484 – .9483 = .0001 in the column of 1â€™.

So we get, Î¸ = 71o 30â€™ + 1â€™ = 71o 31â€™.

(iii) sin Î¸ = .2357

Solution:-

In the table of natural sines, look for a value (â‰¤ .2357) which is sufficiently close to .2357.

We find the value .2351 occurs in the horizontal line beginning with 13o and in the column headed by 36â€™ and in the mean difference, we see .2357 – .2351 = .0006 in the column of 2â€™.

So we get, Î¸ = 13o 36â€™ + 2â€™ = 13o 38â€™.

(iv) sin Î¸ = .6371.

Solution:-

In the table of natural sines, look for a value (â‰¤ .6371) which is sufficiently close to .6371.

We find the value .6361 occurs in the horizontal line beginning with 39o and in the column headed by 30â€™ and in the mean difference, we see .6371 – .6361 = .0010 in the column of 4â€™.

So we get, Î¸ = 39o 30â€™ + 4â€™ = 39o 34â€™.

5. Use the tables to find the acute angle Î¸, given that:Â

(i) cos Î¸ = .4625

Solution:-

In the table of cosines, look for a value (â‰¤ .4625) which is sufficiently close to .4625.

We find the value .4617 occurs in the horizontal line beginning with 62o and in the column headed by 30â€™ and in the mean difference, we see .4625 – .4617 = .0008 in the column of 3â€™.

So we get, Î¸ = 62o 30â€™ – 3â€™ = 62o 27â€™.

(ii) cos Î¸ = .9906Â

Solution:-

In the table of cosines, look for a value (â‰¤ .9906) which is sufficiently close to .9906.

We find the value .9905 occurs in the horizontal line beginning with 7o and in the column headed by 54â€™ and in the mean difference, we see .9906 – .9905 = .0001 in the column of 3â€™.

So we get, Î¸ = 70o 54â€™ – 3â€™ = 70o 51â€™.

(iii) cos Î¸ = .6951

Solution:-

In the table of cosines, look for a value (â‰¤ .6951) which is sufficiently close to .6951.

We find 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 2â€™.

So we get, Î¸ = 46oâ€™ – 2â€™ = 45o 58â€™.

Â

(iv) cos Î¸ = .3412.

Solution:-

In the table of cosines, look for a value (â‰¤ .3412) which is sufficiently close to .3412.

We find the value .3404 occurs in the horizontal line beginning with 70o and in the column headed by 6â€™ and in the mean difference, we see .3412 – .3404 = .0008 in the column of 3â€™.

So we get, Î¸ = 70o 6â€™ – 3â€™ = 70o 3â€™.

6. Use tables to find the acute angle Î¸, given that:Â

(i) tan Î¸ = .2685Â

Solution:-

In the table of natural tangent, look for a value (â‰¤ .2685) which is sufficiently close to .2685.

We find 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 2â€™.

So we get, Î¸ = 15o + 2â€™ = 15o 2â€™.

(ii) tan Î¸ = 1.7451Â

Solution:-

In the table of natural tangent, look for a value (â‰¤ 1.7451) which is sufficiently close to 1.7451.

We find the value 1.7391 occurs in the horizontal line beginning with 60o and in the column headed by 6â€™ and in the mean difference, we see 1.7451 – 1.7391 = .0060 in the column of 5â€™.

So we get, Î¸ = 60o 6â€™ + 5â€™ = 60o 11â€™.

(iii) tan Î¸ = 3.1749Â

Solution:-

In the table of natural tangent, look for a value (â‰¤ 3.1749) which is sufficiently close to 3.1749.

We find the value 3.1716 occurs in the horizontal line beginning with 72o and in the column headed by 30â€™ and in the mean difference, we see 3.1749 – 3.1716 = .0033 in the column of 1â€™.

So we get, Î¸ = 72o 30â€™ + 1â€™ = 72o 31â€™.

(iv) tan Î¸ = .9347

Solution:-

In the table of natural tangent, look for a value (â‰¤ .9347) which is sufficiently close to .9347.

We find 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 4â€™.

So we get, Î¸ = 43o + 4â€™ = 43o 4â€™.

7. Using trigonometric table, find the measure of the angle A when sin A = 0.1822.

Solution:-

In the table of natural sines, look for a value (â‰¤ 0.1822) which is sufficiently close to 0.1822.

We find the value 0.1822 occurs in the horizontal line beginning with 10o and in the column headed by 30â€™.

So we get, A = 10o 30â€™.

8. Using tables, find the value of 2 sin Î¸ â€“ cos Î¸ when (i) Î¸ = 35Â° (ii) tan Î¸ = .2679.

Solution:-

(i) We have to find the value of 2 sin Î¸ â€“ cos Î¸

From the question it is given that, value of Î¸ = 35o

So, substitute the value of Î¸,

= 2 sin 35o â€“ cos 35o

From the table value of sin 35o = .5736 and cos 35o = .8192

= (2 Ã— .5736) – .8192

= 0.3280

(ii) from the question it is given that, tan Î¸ = .2679

In the table of natural sines, look for a value (â‰¤ .2679) which is sufficiently close to .2679.

We find the value column headed by 15o.

So we get, Î¸ = 15o

So, substitute the value of Î¸,

= 2 sin 15o â€“ cos 15o

From the table value of sin 15o = .2588 and cos 15o = .9659

= (2 Ã— .2588) – .9659

= -0.4483

9. If sin xÂ° = 0.67, find the value ofÂ (i) cos xÂ°Â (ii) cos xÂ° + tan xÂ°.

Solution:-

From the question it is given that, sin xo = 0.67.

In the table of natural sines, look for a value (â‰¤ 0.67) which is sufficiently close to 0.67.

We find 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 4â€™.

So we get, Î¸ = 42o + 4â€™ = 42o 4â€™.

Then,

(i) cos xo = cos 42o 4â€²

From the table

= .7431 â€“ .0008Â

= 0.7423 Â

(ii) cos xo + tan xÂ° = cos 42Â° 4â€² + tan 42Â° 4â€²Â

= 0.7423 + .9025Â

= 1.6448

10. If Î¸ is acute and cos Î¸ = .7258, find the value of (i) Î¸ (ii) 2 tan Î¸ â€“ sin Î¸.

Solution:-

From the question, cos Î¸ = .7258Â

In the table of cosines, look for a value (â‰¤ .7258) which is sufficiently close to .7258.

We find the value .7254 occurs in the horizontal line beginning with 43o and in the column headed by 30â€™ and in the mean difference, we see .7258 – .7254 = .0004 in the column of 2â€™.

So we get, Î¸ = 43o 30â€™ – 2â€™ = 43o 28â€™.

(i) Î¸ = 43Â° 30â€² â€“ 2â€™

= 43Â° 28â€².Â

(ii) 2 tan Î¸ â€“ sin Î¸

Substitute the value Î¸,

= 2 tan43Â°28â€² â€“ sin43Â°28â€²Â

= 2 (.9479) â€“ .6879

Â  = 1.8958 â€“ .6879Â

= 1.2079

Therefore, the value of 2 tan Î¸ â€“ sin Î¸ is 1.2079