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}$$\begin{gathered} \sin \beta=\frac{opposite}{hypothenuse} \\ \sin \beta=\frac{14}{16.6433}=0.84\approx0.8(1decimal\text{ place)} \end{gathered}$$\begin{gathered} \cos \beta=\frac{adjacent}{hypothenuse} \\ \cos \beta=\frac{9}{16.6433}=0.5408\approx0.5 \end{gathered}$$\begin{gathered} \tan \beta=\frac{opposite}{adjacent} \\ \tan \beta=\frac{14}{9}=1.555\approx1.6(1d.p) \end{gathered}$$\begin{gathered} \csc \beta=\frac{1}{\sin \beta}=\frac{hypothenuse}{opposite} \\ \csc \beta=\frac{16.6433}{14}=1.1888\approx1.2(1d.p) \end{gathered}$$\begin{gathered} \sec \beta=\frac{1}{\cos \beta}=\frac{hypothenuse}{adjacent} \\ \sec \beta=\frac{16.6433}{9}=1.849\approx1.8(1d.p) \end{gathered}$$\begin{gathered} \cot \beta=\frac{1}{\tan \beta}=\frac{adjacent}{opposite} \\ \cot \beta=\frac{9}{14}=0.643\approx0.6(1d.p) \end{gathered}$