Using tables, find the values of:
(i) sin 83° 12′
(ii) sin 61° 14′
(iii) sin 9° 55′
(iv) sin 32° 40′
Answer
(i) Given,
sin 83° 12′
⇒ sin 83° 12′
From table we have,
∴ sin 83° 12′ = 0.9930
Mean difference of 0′ = 0.0000
Hence, sin 83° 12′ = 0.9930.
(ii) Given,
sin 61° 12′
⇒ sin 61° 12′ + 1'
From table we have,
sin 61° 12′ = 0.8763
Mean difference of 2' = .0003 (To be added)
sin 61° 14′ = 0.8763 + 0.0003 = 0.8766
Hence, sin 61° 14′ = 0.8766.
(iii) Given,
sin 9° 55′
⇒ sin 9° 54′ + 1'
From table we have,
sin 9° 54′ = 0.1719
Mean difference of 1' = .0003 (To be added)
sin 9° 55′ = 0.1719 + 0.0003 = 0.1722
Hence, sin 9° 55′ = 0.1722.
(iv) Given,
sin 32° 40′
⇒ sin 32° 36′ + 4'
From table we have,
sin 32° 36′ = 0.5388
Mean difference of 4' = .0010 (To be added)
sin 32° 40′ = 0.5388 + 0.0010 = 0.5398
Hence, sin 32° 40′ = 0.5398.
Using tables, find the values of:
(i) cos 48° 36′
(ii) cos 23° 6′
(iii) cos 70° 17′
(iv) cos 85° 8′
Answer
(i) Given,
cos 48° 36′
⇒ cos 48° 36′ = cos 48° 36′
From table we have,
cos 48° 36′ = 0.6613
Mean difference of 0′ = 0.0000
Therefore,
cos 48° 36′ = 0.6613
Hence, cos 48° 36′ = 0.6613.
(ii) Given,
cos 23° 6′
⇒ cos 23° 6′ = cos 23° 6′
From table we have,
cos 23° 6′ = 0.9198
Mean difference of 0′ = 0.0000
Therefore,
cos 23° 6′ = 0.9198
Hence, cos 23° 6′ = 0.9198.
(iii) Given,
cos 70° 17′
⇒ cos 70° 18′ − 1′
From table we have,
cos 70° 18′ = 0.3371
Mean difference of 1′ = 0.0003 (to be added)
Therefore,
cos 70° 17′ = 0.3371 + 0.0003
cos 70° 17′ = 0.3374
Hence, cos 70° 17′ = 0.3374.
(iv) Given,
cos 85° 8′
From table we have,
cos 85° 6′ = 0.0854
Mean difference of 2′ = 0.0006 (to be subtracted)
Therefore,
cos 85° 8′ = 0.0854 − 0.0006
cos 85° 8′ = 0.0848
Hence, cos 85° 8′ = 0.0848.
Using tables, find the values of :
(i) tan 24° 24′
(ii) tan 9° 38′
(iii) tan 31° 27′
(iv) tan 65° 50′
Answer
(i) Given,
tan 24° 24′
⇒ tan 24° 24′
From table we have,
tan 24° 24′ = 0.4536
Mean difference of 0′ = 0.0000
Therefore,
tan 24° 24′ = 0.4536
Hence, tan 24° 24′ = 0.4536.
(ii) Given,
tan 9° 38′
⇒ tan 9° 36′ + 2'
From table we have,
tan 9° 36′ = 0.1691
Mean difference of 2′ = 0.0006 (to be added)
Therefore,
tan 9° 38′ = 0.1691 + 0.0006 = 0.1697
Hence, tan 9° 38′ = 0.1697.
(iii) Given,
tan 31° 27′
⇒ tan 31° 24′ + 3'
From table we have,
tan 31° 24′ = 0.6104
Mean difference of 3′ = 0.0012 (to be added)
Therefore,
tan 31° 27′ = 0.6104 + 0.0012 = 0.6116
Hence, tan 31° 27′ = 0.6116.
(iv) Given,
tan 65° 50′
⇒ tan 65° 48′ + 2'
From table we have,
tan 65° 48′ = 2.2251
Mean difference of 2′ = 0.0034 (to be added)
Therefore,
tan 65° 50′ = 2.2251 + 0.0034 = 2.2285
Hence, tan 65° 50′ = 2.2285.
Using tables, find the acute angle θ, when:
sin θ = 0.36
sin θ = 0.4274
sin θ = 0.5955
sin θ = 0.8229
Answer
(i) Given,
sin θ = 0.36
sin 21° 6' = .3600 (From tables)
Difference = .0000
Mean difference for 0' = .0000
Hence, θ = 21° 6'.
(ii) Given,
sin θ = 0.4274
sin 25° 18' = 0.4274 (From tables)
Difference = .0000
Mean difference for 0' = .0000
Hence, θ = 25° 18'.
(iii) Given,
sin θ = 0.5955
sin 36° 30' = 0.5948 (From tables)
Difference = .0007
Mean difference for 3' = .0007
θ = 36° 30'+ 3' = 36° 33'
Hence, θ = 36° 33'.
(iv) Given,
sin θ = 0.8229
sin 55° 18' = 0.8221 (From tables)
Difference = .0008
Mean difference for 5' = .0008
θ = 55° 18'+ 5' = 55° 23'
Hence, θ = 55° 23'.
If sin θ = 0.42, find :
(i) θ
(ii) cos θ
(iii) tan θ
Answer
(i) Given,
sin θ = 0.42
sin 24° 48'= 0.4195 (From tables)
Difference = .0005
Mean difference for 2' = .0005
θ = 24° 48'+ 2' = 24° 50'
Hence, θ = 24° 50'.
(ii) cos θ
cos 24° 48' = 0.9078
Mean difference for 2' = .0002 (to be subtracted)
Therefore,
cos 24° 50' = 0.9078 - 0.0002 = 0.9076
Hence, cos 24° 50' = 0.9076.
(iii) tan θ
tan 24° 48' = 0.4621
Mean difference for 2' = .0007 (to be added)
Therefore,
tan 24° 48' = 0.4621 + 0.0007 = 0.4628
Hence, tan 24° 50' = 0.4628.
Using tables, find the acute angle θ, when:
(i) cos θ = 0.94
(ii) cos θ = 0.8092
(iii) cos θ = 0.1679
Answer
(i) Given,
cos θ = 0.94
cos 19° 54' = .9403
Difference = .0003
Mean difference of 3' = .0003
θ = 19° 54' + 3' = 19° 57'.
Hence, θ = 19° 57'.
(ii) Given,
cos θ = 0.8092
cos 35° 54' = .8100
Difference = .0008
Mean difference of 5' = .0008
θ = 35° 54' + 5' = 35° 59'.
Hence, θ = 35° 59'.
(iii) Given,
cos θ = 0.1679
cos 80° 18' = 0.1685
Difference = .0006
Mean difference of 2' = .0006
θ = 80° 18' + 2' = 80° 20'.
Hence, θ = 80° 20'.
If cos θ = 0.51, find:
(i) θ
(ii) sin θ
(iii)tan θ
Answer
(i) Given,
cos θ = 0.51
cos 59° 18' = .5105
Difference = .0005
Mean difference of 2' = .0005
θ = 59° 18' + 2' = 59° 20'.
Hence, θ = 59° 20'.
(ii) Given,
sin θ
sin 59° 18' = .8599
Mean difference of 2' = .0003
sin 59° 20' = .8599 + .0003 = 0.8602.
Hence, sin 59° 20' = 0.8602.
(iii) Given,
tan θ
tan 59° 18' = 1.6842
Mean difference of 2' = .0022
tan 59° 20' = 1.6842 + .0022= 1.6864.
Hence, tan 59° 20' = 1.6864.
Using tables, find the acute angle θ, when:
(i) tan θ = 1.476
(ii) tan θ = 2.91
(iii) tan θ = 0.3
Answer
(i) Given,
tan θ = 1.476
tan 55° 54' = 1.4770
Difference = .0010
Mean difference of 1' = .0009(to be subtracted)
θ = 55° 54' - 1' = 55° 53'.
Hence, θ = 55° 53'.
(ii) Given,
tan θ = 2.91
tan 71° = 2.9042
Difference = .0058
Mean difference of 2' = .0058(to be added)
θ = 71° + 2' = 71° 2'.
Hence, θ = 71° 2'.
(iii) Given,
tan θ = 0.3
tan 16° 42' = 0.3
Difference = 0.000
Mean difference of 0' = .000
θ = 16° 42'.
Hence, θ = 16° 42'.