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Given 15 cot A = 8, find sin A and sec A. Q.5

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Find the easy way to solve ncert class 10 introduction to trigonometry question of exercise 8.1 question no.5 , its so tricky  for me to solve this question Sir please help me to solve this question in a easy way.Given 15 cot A = 8, find sin A and sec A.


1 Answer

  1. We know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side

    Let us assume a right angled triangle ABC, right angled at B

    sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB

    Let AC be 13k and AB will be 12k

    Where, k is a positive real number.

    According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

    AC2=AB2 + BC2

    Substitute the value of AB and AC

    (13k)2= (12k)2 + BC2

    169k2= 144k2 + BC2

    169k2= 144k2 + BC2

    BC2 = 169k2 – 144k2

    BC2= 25k2

    Therefore, BC = 5k

    Now, substitute the corresponding values in all other trigonometric ratios


    Sin θ = Opposite Side/Hypotenuse = BC/AC = 5/13

    Cos θ = Adjacent Side/Hypotenuse = AB/AC = 12/13

    tan θ = Opposite Side/Adjacent Side = BC/AB = 5/12

    Cosec θ = Hypotenuse/Opposite Side = AC/BC = 13/5

    cot θ = Adjacent Side/Opposite Side = AB/BC = 12/5

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