Question

If O is an angle in standard position and its terminal side passes through the point(9,-7), find the exact value of csc O in simplest radical form.

Answer 1

Let us make a simple diagram to illustrate the question and see the position of the angle

From the right-angled triangle in the diagram, the opposite side is 7, and the adjacent side is 9, we need to find the hypotenuse h.

From Pythagoras theorem,

`\begin{gathered} \text{hyp}^2=opp^2+adj^2 \\ h^2=7^2+9^2 \\ h^2=49+81 \\ h=\sqrt[]{130} \end{gathered}`

Now we know that

`\begin{gathered} \sin \theta=\frac{opposite}{\text{hypotenuse}} \\ \sin \theta=\frac{7}{\sqrt[]{130}} \\ This\text{ position of the angle }\theta\text{ is the 4th quadrant.} \\ In\text{ the 4th quadrant, only cos }\theta\text{ is positive, sin }\theta\text{ and tan }\theta\text{ are negative.} \\ \text{Hence } \\ \sin \theta=-\frac{7}{\sqrt[]{130}} \end{gathered}`

Also

`\begin{gathered} co\sec \theta=(1)/(\sin \theta) \\ co\sec \theta=-\frac{\sqrt[]{130}}{7} \end{gathered}`

Hence, the answer is

`-\frac{\sqrt[]{130}}{7}`