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If sin A = 3/4, Calculate cos A and tan A . Q.3

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Find the best way to solve introduction to trigonometry question of ncert class 10 . What is the easy way to solve the exercise 8.1 question no. 3 , this question is tricky please  help me  to find the simplest way . If sin A = 3/4, Calculate cos A and tan A.

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

    Given: Sin A = 3/4

    We know that, Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side.

    Therefore, Sin A = Opposite side /Hypotenuse= 3/4

    Let BC be 3k and AC will be 4k

    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 AC and BC

    (4k)2=AB2 + (3k)2

    16k2−9k2 =AB2

    AB2=7k2

    Therefore, AB = √7k

    Now, we have to find the value of cos A and tan A

    We know that,

    Cos (A) = Adjacent side/Hypotenuse

    Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get

    AB/AC = √7k/4k = √7/4

    Therefore, cos (A) = √7/4

    tan(A) = Opposite side/Adjacent side

    Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

    BC/AB = 3k/√7k = 3/√7

    Therefore, tan A = 3/√7

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