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TRIGONOMETRY
Trigonometric Tables
Trigonometric Tables The angle for a given trigonometric ratio Solution of a right triangle
Using trigonometric tables, find the values of the following:
1. (i) sin 32 Solution: sin 32 = 0.
(ii) sin 50 Solution: sin 50 = 0.
(iii) sin 43 18 Solution: sin 43 18 = 0.
(iv) sin 72 32 Solution: sin 72 30 = 0. (+) 2 (Mean difference)
(v) sin (12.5) Solution: = sin 12
= sin 12 = x 60 = 30
= sin 12 30
= 0.
(vi) sin (43.1)
Solution: sin (43.1) = sin 43. 1
= sin 43 x 60
= sin 43 6
Using trigonometric tables, find the values of the following:
2. (i) cos 45
Solution : cos 45 = 0.
(ii) cos (48 18 )
Solution: cos 48 18 = 0.6652
(iii) cos (21 24 )
Solution: cos 21 24 = 0.
(iv) cos (48.7)
Solution: cos (48.7) = cos 48 x 60 1 = 60
= cos 48 ( 42 ) = 0.
(v) cos (88 44 )
Solution: cos 88 44 = 0. () 6 (Mean difference)
(vi) cos (26 55 )
Solution: cos 26 54 = 0. () 1
(iii) cosec (20 19 ) Solution: cosec 20 19 =
sin20 18 = 0.
cosec 20 19 = = 2.
(iv) cosec (75 48 ) Solution: cosec 75 48 =
sin 75 48 = 0.
cosec 75 48 = = 1.03156 = 1.
(v) cosec (30.8)
Solution: cosec (30.8) = =
. 8 = x 60 = 48
cosec (30.8) =
sin 30 48 = 0.
cosec (30.8) = = 1.95312 = 1.
(v) cosec (51.4)
Solution: cosec (51.4) = ( x 60 = 24 )
sin20 19
sin 75 48
sin(30.8) sin 30 (. 8 )
sin 30 48
sin 51 (. 4 ) 10
sin 51 24 0.
Using trigonometric tables, find the values of the following:
5. (i) sec 40 Solution sec 40 = =
(ii) sec (40 36 ) Solution: sec 40 36 =
(iii) sec (68 10 ) Solution: sec 68 10 =
cos 68 12 = 0. (5)
sec 68 10 = = 2.
(iv) sec (72 15 )
Solution: sec 72 15 =
cos 72 12 = 0. () 8
sec 72 15 = = 3.
(v) sec (15.3)
Solution: sec (15.3) = sec 15 (. 3)
= sec 15 18 ( . 3 = x 60 = 18 )
cos 40 0.
cos 40 36 1
cos 68 10
cos 72 15
cos15 18 0.
(v) cot (70.5)
Solution: cot (70.5) = =
(vi) cot (15 18 )
Solution: cot 15 18 =
Using trigonometric tables, find the values of the following:
7. (i) tan (51 15 ) + cot (25 18 )
Solution: tan (51 15 ) + cot (25 18 )
= tan 51 15 +
tan 51 12 = 1.
(ii)
Solution: sin 40 + cos20 = 0.6428 + 0.
= 1.
= tan 30 12 = 0.
tan 15 18’
tan (70.5) tan 70 30
1
tan 25 18
sin40 + cos20 tan(30 15 )
sin 40 + cos20 1. tan 30 15 0.
8. Solve the triangle ABC in which A = 25 30 , B = 90 and AB = 10cm.
Solution: A + B + C = 180
25 30 + 90 + C = 180
C = 180 – 25 30 – 90
= 180 – 115 30
C = 64 30
tan A =
tan 25 30 =
10 x tan 25 30 = BC
10 (0.4770) = BC
4.77 = BC
sin 25 30 =
AC =
= 11.08 cm
BC
AB
BC
BC
AC
sin 25 30
A
B 90
10cm^25 ^ 30’
C
10. Find the area of an isosceles triangle with base 10 cm and vertical angle 47 .
Solution: Let AD is the angle bisector of A
AD divides BC equally
DC = BD = 5 cm
tan 23.5 = tan 23 30 =
AD = = = 11.499 = 11.
Area of an isosceles triangle =
= = 57.5 cm^2
11. Find the length of the chord of a circle of radius 6cm, subtending at the centre an
angle of 144 .
Solution:
sin 72 =
x = 6 x sin 72
AB = 2 x = 2(5.706)
= 11.42 cm.
Base x height 2
10 x 11. 5
AD
B tan 23 30 0.
A
x (^) x C
(^47) 23.5
5cm (^) D 5cm 10cm
x 6
O
C B
72
90
6cm
x
144
A C B
O
72
6cm
12. Find the radius of the incircle of a regular polygon of 18 sides each of length 60 cm.
Solution: Angle of a circle = 360
Angle suspended as the centre =
(Here n = 18)
tan 10 =
x =
x = = 170.16 cm
13. Compare the lengths of chords of a circle with radius 3cm subtending angles 108 , 72 at the centre of the circle.
Solution: sin 72 =
x = (sin72) x 3
BC = 2.85 (BC = BM)
AB = 2 x = (2.85) 2
Chord subtending 108 is greater
10 x
n
x
sin 10
10
30 60cm 30
O
x 3
108
A x M x B
O C
72
3cm 72
