Question

An acute angle, θ, is in a right triangle such that cosine of theta equals 15 over 17 period What is the value of csc θ? 17 over 8 17 over 15 8 over 15 8 over 17

Answer 1

By definition we have

`\csc \theta=(1)/(\sin \theta)`

But, remember that

`\sin ^2\theta+\cos ^2\theta=1`

Then we can write sin in function of cos

`\sin ^2\theta+\cos ^2\theta=1\Rightarrow\sin \theta=\sqrt[]{1-\cos ^2\theta}`

Using that we can write csc in function of cos as well

`\csc \theta=\frac{1}{\sqrt[]{1-\cos ^2\theta}}`

We now the value of cos θ here, the let's just put it and do the simplification

`\begin{gathered} \csc \theta=\frac{1}{\sqrt[]{1-\cos^2\theta}} \\ \\ \csc \theta=\frac{1}{\sqrt[]{1-(15^2)/(17^2)}} \\ \\ \csc \theta=\frac{1}{\sqrt[]{1-\frac{225^{}}{289}}} \\ \\ \csc \theta=\frac{1}{\sqrt[]{\frac{289-225^{}}{289}}} \\ \\ \csc \theta=\frac{1}{\sqrt[]{\frac{64^{}}{289}}} \\ \\ \csc \theta=\frac{1}{\frac{8^{}}{17}} \\ \\ \csc \theta=(17)/(8) \end{gathered}`

The correct answer is

`\csc \theta=(17)/(8)`