Answer:
[tex]\begin{gathered} (a)\cot G=\frac{3\sqrt{7}}{7} \\ (b)\sin E=\frac{3}{4} \\ (c)\sec G=\frac{4}{3} \end{gathered}[/tex]
Explanation:
A sketch of the triangle and the given dimensions is attached below:
First, find the length of EF using the Pythagorean Theorem:
[tex]\begin{gathered} EG^2=EF^2+FG^2 \\ 8^2=EF^2+6^2 \\ EF^2=8^2-6^2=64-36=28 \\ EF=\sqrt{28}=2\sqrt{7} \end{gathered}[/tex]
(a)cot G
Cotangent is the inverse of tangent.
• The side length opposite to G = 2√7
,
• The side length adjacent to G = 6
[tex]\begin{gathered} \cot G=\frac{Adjacent}{Opposite}=\frac{6}{2\sqrt{7}}=\frac{3}{\sqrt{7}} \\ Rationalising: \\ \cot G=\frac{3}{\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}} \\ \implies\cot G=\frac{3\sqrt{7}}{7} \end{gathered}[/tex]
(b)sin E
• The side length ,opposite, to E = 6
,
• The length of the ,hypotenuse, = 8
[tex]\begin{gathered} \sin E=\frac{Opposite}{Hypotenuse}=\frac{6}{8} \\ \implies\sin E=\frac{3}{4} \end{gathered}[/tex]
(c)sec G
Secant is the inverse of cosine.
• The side length adjacent to G = 6
,
• The length of the hypotenuse = 8
[tex]\begin{gathered} \sec G=\frac{Hypotenuse}{Adjacent}=\frac{8}{6} \\ \implies\sec G=\frac{4}{3} \end{gathered}[/tex]
