Example 1 :
From the given figure, find all the trigonometric ratios of angle B.
Solution :
Here we have to mark the angle at B.
From the given triangle,
Hypotenuse side (BC) = 41
Opposite side (AC) = 9
Adjacent side (AB) = 40
sin B = Opposite side / Hypotenuse side
sin B = AC/BC = 6/41
cos B = Adjacent side / Hypotenuse side
cos B = AB/BC = 40/41
tan B = Opposite side / Adjacent side
tan B = AC/AB = 6/40
cosec B = Hypotenuse side/Opposite side
cosec B = BC/AC = 41/6
sec B = Hypotenuse side/Adjacent side
sec B = BC/AB = 41/40
cot B = Adjacent side / Opposite side
cot B = AB/AC = 40/6
Example 2 :
From the given figure, find the values of
(i) sin B (ii) sec B (iii) cot B
(iv) cos C (v) tan C (vi) cosec C
Solution :
In triangle ABD,
AB2 = AD2 + BD2
132 = AD2 + 52
169 - 25 = AD2
AD2 = 144
AD = 12
In triangle ADC,
AC2 = AD2 + DC2
AC2 = 122 + 162
AC2 = 144 + 256
AC = 20
(i) sin B = AD/AB sin B = 12/13 (ii) sec B = AB/BD sec B = 13/5 (iii) cot B = BD/AD cot B = 5/12 |
(iv) cos C = DC/AC cos C = 16/20 = 4/5 (v) tan C = AD/DC tan C = 12/16 = 3/4 (vi) cosec C = AC/AD cosec C = 20/12 = 5/3 |
Example 3 :
If 2 cos θ = √3, then find all the trigonometric ratios of angle θ.
Solution :
cos θ = √3/2 = Adjacent side / Hypotenuse side
(Opposite side)2 = (Hypotenuse side)2 - (Adjacent side)2
(Opposite side)2 = 22 - (√3)2 = 4 - 3
Opposite side = 1
sin A = Opposite side / Hypotenuse = 1/2
cos A = Adjacent side / Hypotenuse = √3/2
tan A = Opposite side / Adjacent side = 1/√3
cosec A = Hypotenuse / Opposite side = 2/1
sec A = Hypotenuse / Adjacent side = 2/√3
cot A = Adjacent side / Opposite side = √3/1
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