We have, tanθ = 15k/8k = perpendicular/base (For some value of k)

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

∴AB² = BC² + AC²

AB² = (15k)² + (8k)²

AB² = 225k² + 64k²

AB² = 289k² = (17k)²

→ AB = 17k

Hence, the trignometeric ratios for the given θ are:

sinθ = BC/AB = (15k)/(17k) = 15/17

cosθ = AC/AB = (8k)/(17k) = 8/17

tanθ = 15/8

cotθ = AC/BC = 1/tanθ = 8/15

cosecθ = AB/BC = 1/sinθ = 17/15

secθ = AB/AC = 1/cosθ = 17/8