Answer :

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

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

∴AB^{2} = BC^{2} + AC^{2}

AB^{2} = (15k)^{2} + (8k)^{2}

AB^{2} = 225k^{2} + 64k^{2}

AB^{2} = 289k^{2} = (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

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