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.
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′
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