Find sec θ if tan θ = 15/8 and θ terminates in QIII.

1 answer:

5 0

Tangent = opposite/adjacent
tangent = 15 / 8
If opp = 15 and adj = 8 then
hypotenuse^2 = 15^2 + 8^2
hypotenuse^2 = 225 + 64
hypotenuse^2 = 289
hypotenuse = 17

secant = hypotenuse / adjacent
secant = 17 / 8
secant = 2.125

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