Answer:
[tex]\cos G=\dfrac{2}{3}[/tex]
[tex]\csc E=\dfrac{3}{2}[/tex]
[tex]\cot G=\dfrac{2}{\sqrt{5}}[/tex]
Step-by-step explanation:
If the right angle of right triangle EFG is ∠F, then EG is the hypotenuse, and EF and FG are the legs of the triangle. (Refer to attached diagram).
Given ΔEFG is a right triangle, and EG = 6 and FG = 4, we can use Pythagoras Theorem to calculate the length of EF.
[tex]\begin{aligned}EF^2+FG^2&=EG^2\\EF^2+4^2&=6^2\\EF^2+16&=36\\EF^2&=20\\\sqrt{EF^2}&=\sqrt{20}\\EF&=2\sqrt{5}\end{aligned}[/tex]
Therefore:
[tex]\hrulefill[/tex]
To find cos G, use the cosine trigonometric ratio:
[tex]\boxed{\begin{minipage}{9 cm}\underline{Cosine trigonometric ratio} \\\\$\sf \cos(\theta)=\dfrac{A}{H}$\\\\where:\\ \phantom{ww}$\bullet$ $\theta$ is the angle. \\ \phantom{ww}$\bullet$ $\sf A$ is the side adjacent the angle. \\\phantom{ww}$\bullet$ $\sf H$ is the hypotenuse (the side opposite the right angle). \\\end{minipage}}[/tex]
For angle G, the adjacent side is FG and the hypotenuse is EG.
Therefore:
[tex]\cos G=\dfrac{FG}{EG}=\dfrac{4}{6}=\dfrac{2}{3}[/tex]
[tex]\hrulefill[/tex]
To find csc E, use the cosecant trigonometric ratio:
[tex]\boxed{\begin{minipage}{9 cm}\underline{Cosecant trigonometric ratio} \\\\$\sf \csc(\theta)=\dfrac{H}{O}$\\\\where:\\ \phantom{ww}$\bullet$ $\theta$ is the angle. \\ \phantom{ww}$\bullet$ $\sf A$ is the side adjacent the angle. \\\phantom{ww}$\bullet$ $\sf H$ is the hypotenuse (the side opposite the right angle). \\\end{minipage}}[/tex]
For angle E, the hypotenuse is EG and the opposite side is FG.
Therefore:
[tex]\csc E=\dfrac{EG}{FG}=\dfrac{6}{4}=\dfrac{3}{2}[/tex]
[tex]\hrulefill[/tex]
To find cot G, use the cotangent trigonometric ratio:
[tex]\boxed{\begin{minipage}{9 cm}\underline{Cotangent trigonometric ratio} \\\\$\sf \cot(\theta)=\dfrac{A}{O}$\\\\where:\\ \phantom{ww}$\bullet$ $\theta$ is the angle. \\ \phantom{ww}$\bullet$ $\sf A$ is the side adjacent the angle. \\\phantom{ww}$\bullet$ $\sf H$ is the hypotenuse (the side opposite the right angle). \\\end{minipage}}[/tex]
For angle G, the adjacent side is FG and the opposite side is EF.
Therefore:
[tex]\cot G=\dfrac{FG}{EF}=\dfrac{4}{2\sqrt{5}}=\dfrac{2}{\sqrt{5}}[/tex]
