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Chapter 19

Trigonometrical Tables

Class - 10 ML Aggarwal Understanding ICSE Mathematics



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'

Answer

(i)

sin 35° 22' = sin (35° 18' + 4')
sin 35° 18' = .5779
Mean difference for 4' = .0010       (To be added)
sin 35° 22' = .5779 + .0010 = .5789

Hence, the value of sin 35° 22' is .5789

(ii)

sin 71° 31' = sin (71° 30' + 1')
sin 71° 30' = .9483
Mean difference for 1' = .0001       (To be added)
sin 71° 31' = .9483 + .0001 = .9484

Hence, the value of sin 71° 31' is .9484

(iii)

sin 65° 20' = sin (65° 18' + 2')
sin 65° 18' = .9085
Mean difference for 2' = .0002       (To be added)
sin 65° 20' = .9085 + .0002 = .9087

Hence, the value of sin 65° 20' is .9087

(iv)

sin 23° 56' = sin (23° 54' + 2')
sin 23° 54' = 0.4051
Mean difference of 2' = .0005       (To be added)
sin 23° 56' = .4051 + .0005 = .4056

Hence, the value of sin 23° 56' is .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'

Answer

(i)

cos 62° 27' = cos (62° 24' + 3')
cos 62° 24' = 0.4633
Mean difference of 3' = .0008       (To be subtracted)
cos 62° 27' = .4633 - .0008 = .4625

Hence, the value of cos 62° 27' = .4625

(ii)

cos 3° 11' = cos (3° 6' + 5')
cos 3° 6' = .9985
Mean difference of 5' = .0001       (To be subtracted)
cos 3° 11' = .9985 - .0001 = .9984

Hence, the value of cos 3° 11' = .9984

(iii)

cos 86° 40' = cos (86° 36' + 4')
cos 86° 36' = .0593
Mean difference of 4' = .0012       (To be subtracted)
cos 86° 40' = .0593 - .0012 = .0581

Hence, the value of cos 86° 40' = .0581

(iv)

cos 45° 58' = cos (45° 54' + 4')
cos 45° 54' = .6959
Mean difference of 4' = .0008       (To be subtracted)
cos 45° 58' = .6959 - .0008 = .6951

Hence, the value of cos 45° 58' = .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'

Answer

(i)

tan 15° 2' = tan(15° 0' + 2')
tan 15° 0' = .2679
Mean difference of 2' = .0006       (To be added)
tan 15° 2' = .2679 + .0006 = .2685

Hence, the value of cos 15° 2' = .2685

(ii)

tan 53° 14' = tan(53° 12' + 2')
tan 53° 12' = 1.3367
Mean difference of 2' = .0016       (To be added)
tan 53° 12' = 1.3367 + .0016 = 1.3383

Hence, the value of cos 53° 12' = 1.3383

(iii)

tan 82° 18' = 7.3962 (directly from tables)

Hence, the value of tan 82° 18' = 7.3962

(iv)

tan 6° 9' = tan (6° 6' + 3')
tan 6° 6' = .1069
Mean difference of 3' = .0009       (To be added)
tan 6° 9' = .1069 + .0009 = .1078

Hence, the value of tan 6° 9' = 0.1078

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

Answer

(i) Given,

sin θ = .5789
sin 35° 18' = .5779       (From tables)
Difference = .0010
Mean difference for 4' = .0010
θ = 35° 18' + 4' = 35° 22'.

Hence, the value of θ = 35° 22'.

(ii) Given,

sin θ = .9484
sin 71° 30' = .9483       (From tables)
Difference = .0001
Mean difference for 1' = .0001
θ = 71° 30' + 1' = 71° 31'.

Hence, the value of θ = 71° 31'.

(iii) Given,

sin θ = .2357
sin 13° 36' = .2351       (From tables)
Difference = .0006
Mean difference for 2' = .0006
θ = 13° 36' + 2' = 13° 38'.

Hence, the value of θ = 13° 38'.

(iv) Given,

sin θ = .6371
sin 39° 30' = .6361       (From tables)
Difference = .0010

Mean difference for 4' = .0009
and
Mean difference for 5' = .0011

Taking Mean difference for 4' = .0009
θ = 39° 30' + 4' = 39° 34'.

Hence, the value of θ = 39° 34'.

Question 5

Use 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

Answer

(i) Given,

cos θ = .4625
cos 62° 30' = .4617       (From tables)
Difference = .0008
Mean difference for 3' = .0008
θ = 62° 30' - 3' = 62° 27'.

Hence, the value of θ = 62° 27'.

(ii) Given,

cos θ = .9906
cos 7° 54' = .9905       (From tables)
Difference = .0001
Mean difference for 3' = .0001
θ = 7° 54' - 3' = 7° 51'.

Hence, the value of θ = 7° 51'.

(iii) Given,

cos θ = .6951
cos 46° 0' = .6947       (From tables)
Difference = .0004
Mean difference for 2' = .0004
θ = 46° 0' - 2' = 45° 58'.

Hence, the value of θ = 45° 58'.

(iv) Given,

cos θ = .3412
cos 70° 6' = .3404       (From tables)
Difference = .0008
Mean difference for 3' = .0008
θ = 70° 6' - 3' = 70° 3'.

Hence, the value of θ = 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

Answer

(i) Given,

tan θ = .2685
tan 15° 0' = .2679       (From tables)
Difference = .0006
Mean difference for 2' = .0006
θ = 15° 0' + 2' = 15° 2'

Hence, the value of θ = 15° 2'.

(ii) Given,

tan θ = 1.7451
tan 60° 6' = 1.7391       (From tables)
Difference = .0060
Mean difference for 5' = .0060
θ = 60° 6' + 5' = 60° 11'

Hence, the value of θ = 60° 11'.

(iii) Given,

tan θ = 3.1749
tan 72° 30' = 3.1716       (From tables)
Difference = .0033
Mean difference for 1' = .0033
θ = 72° 30' + 1' = 72° 31'

Hence, the value of θ = 72° 31'.

(iv) Given,

tan θ = .9347
tan 43° 0' = .9325       (From tables)
Difference = .0022
Mean difference for 4' = .0022
θ = 43° 0' + 4' = 43° 4'

Hence, the value of θ = 43° 4'.

Question 7

Using trigonometric table, find the measure of the angle A when sin A = 0.1822.

Answer

sin A = 0.1822
sin 10° 30' = 0.1822       (From tables)
A = 10° 30'

Hence, the value of A = 10° 30'.

Question 8

Using tables, find the value of 2 sin θ - cos θ when

(i) θ = 35°

(ii) tan θ = .2679.

Answer

(i) Putting θ = 35° in 2 sin θ - cos θ,

⇒ 2 sin 35° - cos 35°
⇒ 2 × .5736 - .8192
⇒ 1.1472 - .8192
⇒ .3280

Hence, the value of 2 sin θ - cos θ = .3280

(ii) Given,

tan θ = .2679
tan θ = tan 15°
θ = 15°.

Putting θ = 15° in 2 sin θ - cos θ,

⇒ 2 sin 15° - cos 15°
⇒ 2 × .2588 - .9659
⇒ .5176 - .9659
⇒ -.4483

Hence, the value of 2 sin θ - cos θ = -0.4483

Question 9

If sin x° = 0.67, find the value of

(i) cos x°

(ii) cos x° + tan x°

Answer

(i) Given,

sin x° = 0.67
sin 42° = .6691       (From tables)
Difference = .0009
Mean difference for 4' = .0009
x° = 42° + 4' = 42° 4'.

Finding cos x°,

cos 42° = .7431
Mean difference of 4' = .0008
cos 42° 4' = .7431 - .0008 = .7423

Hence, the value of cos 42° 4' = .7423

(ii) Calculating tan x°

tan 42° = .9004
Mean difference of 4' = .0021
tan 42° 4' = .9004 + .0021 = .9025

⇒ cos x° + tan x° = .7423 + .9025 = 1.6448

Hence, the value of cos x° + tan x° = 1.6448.

Question 10

If θ is acute and cos θ = .7258, find the value of

(i) θ

(ii) 2 tan θ - sin θ

Answer

(i) Given,

cos θ = .7258
cos 43° 30' = .7254
Difference = .0004
Mean difference of 2' = .0004
θ = 43° 30' - 2' = 43° 28'.

Hence, the value of θ = 43° 28'.

(ii) Putting, θ = 43° 28' in 2 tan θ - sin θ,

Finding value of tan 43° 28',

tan 43° 24' = .9457
Mean difference of 4' = .0022
tan 43° 28' = .9457 + .0022 = .9479

Finding value of sin 43° 28',

sin 43° 24' = .6871
Mean difference of 4' = .0008
sin 43° 28' = .6871 + .0008 = .6879

2 tan θ - sin θ,

⇒ 2 × .9479 - .6879
⇒ 1.8958 - .6879
⇒ 1.2079

Hence, the value of 2 tan θ - sin θ = 1.2079

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