If sin A = 3/4, Calculate cos A and tan A.find it asap

Dear Student

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,

AC^2=AB^^2+ BC2
Substitute the value of AC and BC
(4k)^2=AB^2+ (3k)^2
16k^2−9k^2=AB^2
AB^2=7k^2
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 getAB/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

Thanks