Let ΔABC be a right-angled triangle, right-angled at point B

Given that,

Sin A =

Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

(4k)² = AB² + (3k)²

16k² - 9k² = AB²

7k² = AB²

AB = k

Cos A =

= =

Tan A =

= =