Question

Sin 0 =(8)/(9), tan 0 >0 find sec 0

Answer 1

Answer:

$\displaystyle \sec(\theta)=\frac{9\sqrt{17}}{17}$

Step-by-step explanation:

We are given that:

$\displaystyle \sin(\theta)=\frac{8}{9}\text{ and } \tan(\theta)>0$

And we want to find sec(θ).

First, note that both sine and tangent are positive. The only quadrant in which this can occur is QI. Hence, all trig ratios will be positive.

Also, recall that sine is the ratio of the opposite side to the hypotenuse. Using this information and the Pythagorean Theorem, we can determine the adjacent side:

$a=\sqrt{9^2-8^2}=\sqrt{17}$

So, with respect to θ, the adjacent side is √(17), the opposite side is 8, and the hypotenuse is 9.

Secant is the ratio of the hypotenuse to the adjacent. Hence:

$\displaystyle \sec(\theta)=\frac{9}{\sqrt{17}}=\frac{9\sqrt{17}}{17}$

Again, since θ is in QI, all trig ratios are positive.