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 35 ^{o} 22â€²**

**Solution:-**

To find the value of sin 35^{o} 22â€™,

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

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 35^{o} 22â€™ is obtained as under,

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

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

Then, sin 35^{o} 22â€™ = 0.5789

(ii) sin 71^{o} 31â€²

**Solution:-**

To find the value of sin 71^{o} 31â€™,

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

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 71^{o} 31â€™ is obtained as under,

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

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

Then, sin 71^{o} 31â€™ = 0.9484

(iii) sin 65^{o} 20â€²

**Solution:-**

To find the value of sin 65^{o} 20â€™,

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

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 65^{o} 20â€™ is obtained as under,

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

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

Then, sin 65^{o} 20â€™ = 0.9087

(iv) sin 23^{o} 56â€²

**Solution:-**

To find the value of sin 23^{o} 56â€™,

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

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 23^{o} 56â€™ is obtained as under,

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

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

Then, sin 23^{o} 56â€™ = 0.4056

**2. Find the value of the following:(i) cos 62 ^{o} 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 62^{o} 27â€™,

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

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 62^{o} 27â€™ is 0.4625.

(ii) cos 3^{o} 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 3^{o} 11â€²,

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

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 3^{o} 11â€² is 0.9984.

(iii) cos 86^{o} 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 86^{o} 40â€²,

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

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 86^{o} 40â€² is 0.0581.

(iv) cos 45^{o} 58â€².

**Solution:-**

To find the value of cos 45^{o} 58â€²,

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

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 45^{o} 58â€² is 0.6951.

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

**Solution:-**

To find the value of tan 15^{o} 2â€²,

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

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 15^{o} 2â€™ is 0.2685.

(ii) tan 53^{o} 14â€²

**Solution:-**

To find the value of tan 53^{o} 14â€²,

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

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 53^{o} 14â€² is 1.3383.

(iii) tan 82^{o} 18â€²

**Solution:-**

To find the value of tan 82^{o} 18â€²,

We read the table of natural sines in the horizontal line which begins with 82^{o}

Then, value that we find in vertical column is 7.3962

Therefore, tan 82^{o} 18â€™ is 7.3962

(iv) tan 6^{o} 9â€².

**Solution:-**

To find the value of tan 6^{o} 9â€²,

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

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 6^{o} 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 35^{o} 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¸ = 35^{o} 18â€™ + 4â€™ = 35^{o} 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 71^{o} 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¸ = 71^{o} 30â€™ + 1â€™ = 71^{o} 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 13^{o} 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¸ = 13^{o} 36â€™ + 2â€™ = 13^{o} 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 39^{o} 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¸ = 39^{o} 30â€™ + 4â€™ = 39^{o} 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 62^{o} 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¸ = 62^{o} 30â€™ – 3â€™ = 62^{o} 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 7^{o} 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¸ = 70^{o} 54â€™ – 3â€™ = 70^{o} 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 46^{o} and in the mean difference, we see .6951 – .6947 = .0004 in the column of 2â€™.

So we get, Î¸ = 46^{o}â€™ – 2â€™ = 45^{o} 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 70^{o} 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¸ = 70^{o} 6â€™ – 3â€™ = 70^{o} 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 15^{o} and in the mean difference, we see .2685 – .2679 = .0006 in the column of 2â€™.

So we get, Î¸ = 15^{o} + 2â€™ = 15^{o} 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 60^{o} 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¸ = 60^{o} 6â€™ + 5â€™ = 60^{o} 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 72^{o} 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¸ = 72^{o} 30â€™ + 1â€™ = 72^{o} 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 43^{o} and in the mean difference, we see .9347 – .9325 = .0022 in the column of 4â€™.

So we get, Î¸ = 43^{o} + 4â€™ = 43^{o} 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 10^{o} and in the column headed by 30â€™.

So we get, A = 10^{o} 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¸ = 35^{o}

So, substitute the value of Î¸,

= 2 sin 35^{o} â€“ cos 35^{o}

From the table value of sin 35^{o} = .5736 and cos 35^{o} = .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 15^{o}.

So we get, Î¸ = 15^{o}

So, substitute the value of Î¸,

= 2 sin 15^{o} â€“ cos 15^{o}

From the table value of sin 15^{o} = .2588 and cos 15^{o} = .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 x^{o} = 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 42^{o} and in the mean difference, we see 0.6700 – 0.6691 = .0009 in the column of 4â€™.

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

Then,

(i) cos x^{o} = cos 42^{o} 4â€²

From the table

= .7431 â€“ .0008Â

= 0.7423 Â

(ii) cos x^{o} + 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 43^{o} 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¸ = 43^{o} 30â€™ – 2â€™ = 43^{o} 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