Right triangle EFG has its right angle at F, EG = 6 , and FG = 4 What is the value of the trigonometric ratio of an angle of the triangle? Drag a value to each box to match the trigonometric ratio with its value .​

Right triangle EFG has its right angle at F EG 6 and FG 4 What is the value of the trigonometric ratio of an angle of the triangle Drag a value to each box to m class=


Answer :

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:

  • EF = 2√5
  • FG = 4
  • EG = 6

[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]

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