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
