We have, sinA = (9k)/(41k) = BC/AB (For some value of k)

By Pythagoras theorem, (hypotenuse)² = (perpendicular)² + (base)²

∴AB² = BC² + AC²

= (41k)² = (9k)² + AC²

= 1681k² = 81k² + AC²

AC² = 1600k²

= (40k)²

→ AC = 40k

∴ cosA = AC/AB = (40k)/(41k) = 40/41

tanA = BC/AC = sinA/cosA = 9/40