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If sin A = 3/4, Calculate cos A and tan A. find it asap
Dear StudentLet us assume a right angled triangle ABC, right angled at BGiven: Sin A = 3/4We 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= 3/4Let BC be 3k and AC will be 4kwhere k is a positive real number.According to the Pythagoras theorem,AC^2=AB^^2+ BC2Substitute the value of AC and BC(4k)^2=AB^2+ (3k)^216k^2−9k^2=AB^2AB^2=7k^2Therefore, AB =√7kNow, we have to find the value of cos A and tan AWe know that,Cos (A) =Adjacent side/HypotenuseSubstitute the value of AB and AC and cancel the constant k in both numerator and denominator, we getAB/AC = √7k/4k = √7/4Therefore, cos (A)= √7/4tan(A) = Opposite side/Adjacent sideSubstitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,BC/AB = 3k/√7k = 3/√7Therefore, tan A= 3/√7Thanks
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