Question

Find cos θ if sin θ = -(5/13) and tan θ > 0

Answer 1

$$\bf sin(\theta)= -\cfrac{5}{13} \implies \cfrac{opposite}{hypotenuse} \implies \frac{b}{c}\\
c^2=a^2+b^2 \implies b= \pm\sqrt{c^2-a^2}
\implies b= \pm\sqrt{144} \implies b=12\\
\text{now } tan(\theta) > 0 \text{ meaning is positive}
$$
so for \( tan(\theta) to be positive it has to have either sine or cosine negative and the other positve
as we can see above, sine is already negative \(\bf in(\theta)= -\cfrac{5}{13}
so to get a positive tangent, cosine has to also be negative, thus
b = -12

Answer 2

hmm, lemme redo that :/

Answer 3

$$\bf
sin(\theta)= -\cfrac{5}{13} \implies \cfrac{opposite}{hypotenuse} \implies \frac{b}{c}\\
c^2=a^2+b^2 \implies b= \pm\sqrt{c^2-a^2}
\implies b= \pm\sqrt{144} \implies b=12\\
\text{now } tan(\theta) > 0 \text{ meaning is positive}
$$
so for \( tan(\theta)\) to be positive it has to have either sine or cosine negative and the other positve
as we can see above, sine is already negative, \(\bf sin(\theta)= -\cfrac{5}{13}\)
so to get a positive tangent, cosine has to also be negative, thus
b = -12

Answer 4

Haha i understood what you were saying :P no need to redo. and that's what i thought. so it would be -12/13 correct?

Answer 5

wait a second, I'm after "a", NOT b component

Answer 6

oh :P

Answer 7

well, the same value will come out anyway :)

Answer 8

\(\bf sin(\theta)= -\cfrac{5}{13} \implies \cfrac{opposite}{hypotenuse} \implies \frac{b}{c}\\
c^2=a^2+b^2 \implies a= \pm\sqrt{c^2-b^2}
\implies a= \pm\sqrt{144} \implies a=12\)

Answer 9

So it should be -12/13 correct?

Answer 10

and it will be -12 because the tangent is positive, so the adjacent side will also be negative

Answer 11

yes, is correct

Answer 12

thats what i get though?