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Question:

If sinA = 3/4, calculate cosA and tanA.

Solution:

Let ABC is right △, right angled at B. sinA = 3/4 = BC/AC. Let BC = 3x and AC = 4x. (x +ve direction) AC² = AB² + BC² ⇒ AB² = (4x)² - (3x)² = 7x² ⇒ AB = √7x ⇒ cosA = AB/AC = √7k/4k = √7/4 ⇒ tanA = BC/AB = 3/√7