# Math Help - Finding sin and tan from secant?

1. ## Finding sin and tan from secant?

If secθ = 17, what is sinθ? tanθ?

In order to do this would we have to draw a triangle? Secant is hyp/opp. Thus should the hyp be 17 and the opp be 1?

If that is the case will the answer be sinθ = 16.97/17 and tanθ = 16.97/1 = 16.97?

PS: I got the 16.97 from using the a^2 + b^2 = c^2 formula. Please tell me if this is correct or is there something else to it. Thank you so much.

2. Originally Posted by nox16
If secθ = 17, what is sinθ? tanθ?

In order to do this would we have to draw a triangle? Secant is hyp/opp. Thus should the hyp be 17 and the opp be 1?

If that is the case will the answer be sinθ = 16.97/17 and tanθ = 16.97/1 = 16.97?

PS: I got the 16.97 from using the a^2 + b^2 = c^2 formula. Please tell me if this is correct or is there something else to it. Thank you so much.
$sec\theta = \frac{\text{hypotenuse(h)}}{\text{base(b)}} = \frac{17}{1}$

$\therefore$ by Pythagorean theorem:

$h^2= p^2+ b^2$

find out what the perpendicular would be!

then, use:

$sin\theta = \frac{\text{perpendicular(p)}}{\text{hypotenuse(h) }}$

and

$tan\theta =\frac{\text{perpendicular(p)}}{\text{base(b)}}$

3. I think that what I am doing. I realized I made a mistake typing out the secantθ formula. It is actually adjacent/hyp.

4. Originally Posted by nox16
I think that what I am doing. I realized I made a mistake typing out the secantθ formula. It is actually adjacent/hyp.
wait..

adjacent/hyp = cosine(theta). secant is its reciprocal.

$sec\theta = \frac{\text{hypotenuse(h)}}{\text{adjacent(b)}}$

See this if you are not clear yet:

Trigonometric functions - Wikipedia, the free encyclopedia

5. Thanks for your help nonetheless.

*Dang keep on messing up on these trig functions. I got it correctly on paper though.