Given that cos θ = –12∕13 and θ is in quadrant III, find csc θ.

Mathematics

2 answers:

Zanzabum2 years ago

8 0

Answer:

Step-by-step explanation:

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Trava [24]2 years ago

7 0

Answer:

Step-by-step explanation:

We are given:

$\displaystyle \cos(\theta)=-\frac{12}{13},\,\theta\in\text{QIII}$

Since cosine is the ratio of the adjacent side over the hypotenuse, this means that the opposite side is (we can ignore negatives for now):

$o=\sqrt{13^2-12^2}=\sqrt{25}=5$

So, the opposite side is 5, the adjacent side is 12, and the hypotenuse is 13.

And since θ is in QIII, sine/cosecant is negative, cosine/secant is negative, and tangent/cotangent is positive.

Cosecant is given by the hypotenuse over the opposite side. Thus:

$\displaystyle \csc(\theta)=\frac{13}{5}$

Since θ is in QIII, cosecant must be negative:

$\displaystyle \csc(\theta)=-\frac{13}{5}$

Our answer is A.

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Answer:

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