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

Question 20 options:

A)

–13∕5


B)

13∕5

C)

5∕13

D)

–5∕13

User Whelch
by
4.7k points

2 Answers

2 votes

Answer:

A

Explanation:

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User Marc Polizzi
by
4.1k points
9 votes

Answer:

A

Explanation:

We are given:


\displaystyle \cos(\theta)=-(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=√(13^2-12^2)=√(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)=(13)/(5)

Since θ is in QIII, cosecant must be negative:


\displaystyle \csc(\theta)=-(13)/(5)

Our answer is A.

User Christian Hagelid
by
4.7k points