# FINDING TRIGONOMETRIC RATIOS FROM THE GIVEN TRIANGLE

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

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,

AC2  =  122 + 162

AC2  =  144 + 256

AC  =  20

 (i) sin B  = AD/ABsin B  =  12/13(ii) sec B = AB/BDsec B  =  13/5 (iii) cot B  =  BD/ADcot B  =  5/12 (iv) cos C  = DC/ACcos C  =  16/20 = 4/5(v) tan C  =  AD/DCtan C  =  12/16 = 3/4(vi) cosec C  =  AC/ADcosec 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)  =  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 Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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