Given that: sin A = 3/4
Let sin A = 3k/4/, where k is a real number.
In ∆ABC, by Pythagoras theorem, we have
AB² = AC²-BC²
= (4k)²-(3k)²
= 16k²-9k²
= 7k²
⇒ AB = √(7k²) = √7k
Hence, cos A = AB/AC = √7k/4k = √7/4 and tan A = BC/AB = 3k/√7k = 3/√7

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