Question

Please helppppp this is super hard and I need some help. Thank you:)

Answer 1

Given that the terminal side of an angles passesthrough the point (9, -14)

This can be represented diagramatically as shown below

From the diagram above, calculate x using the pythagoras theorem

`\begin{gathered} x^2=9^2+14^2 \\ x^2=81+196 \\ x^2=277 \\ x=\sqrt[]{277} \\ x=16.6433 \end{gathered}`
`\begin{gathered} \sin \beta=(opposite)/(hypothenuse) \\ \sin \beta=(14)/(16.6433)=0.84\approx0.8(1decimal\text{ place)} \end{gathered}`
`\begin{gathered} \cos \beta=(adjacent)/(hypothenuse) \\ \cos \beta=(9)/(16.6433)=0.5408\approx0.5 \end{gathered}`
`\begin{gathered} \tan \beta=(opposite)/(adjacent) \\ \tan \beta=(14)/(9)=1.555\approx1.6(1d.p) \end{gathered}`
`\begin{gathered} \csc \beta=(1)/(\sin \beta)=(hypothenuse)/(opposite) \\ \csc \beta=(16.6433)/(14)=1.1888\approx1.2(1d.p) \end{gathered}`
`\begin{gathered} \sec \beta=(1)/(\cos \beta)=(hypothenuse)/(adjacent) \\ \sec \beta=(16.6433)/(9)=1.849\approx1.8(1d.p) \end{gathered}`
`\begin{gathered} \cot \beta=(1)/(\tan \beta)=(adjacent)/(opposite) \\ \cot \beta=(9)/(14)=0.643\approx0.6(1d.p) \end{gathered}`