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

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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

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