ML Aggarwal Solution Class 10 Chapter 19 Trigonometric Tables Exercise 19

 Exercise 19

Question 1

Find the value of the following:

(i) sin 35° 22′

(ii) sin 71° 31′

(iii) sin 65° 20′

(iv) sin 23° 56′.

Sol :

(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′.

Sol :

(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′.

Sol :

(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.

Sol :

(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.

Sol :

(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

Sol :

(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.

Sol :

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°.

Sol :

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.

Sol :

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