Respuesta :

tramserran

Answer:  [tex]\bold{5)\ \cos \theta=\dfrac{\sqrt{11}}{6}\qquad 6)\ \tan \theta =\dfrac{8}{17}\qquad 7)\ \cos \theta = \dfrac{4}{3}\qquad 8)\ \cos \theta = \dfrac{\sqrt{10}}{10}}[/tex]

Step-by-Step Explanation:

Pythagorean Theorem is: a² + b² = c²  , where "c" is the hypotenuse

[tex]5)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{3\sqrt{11}}{18}\quad \rightarrow \large\boxed{\dfrac{\sqrt{11}}{6}}[/tex]

Note: (15)² + (3√11)² = hypotenuse²   →   hypotenuse = 18

[tex]6)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{8}{17}\quad =\large\boxed{\dfrac{8}{17}}[/tex]

Note: 8² + 15² = hypotenuse²   →   hypotenuse = 17

[tex]7)\ \tan \theta=\dfrac{\text{side opposite to}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{20}{15}\quad \rightarrow \large\boxed{\dfrac{4}{3}}[/tex]

Note: hypotenuse not needed for tan

[tex]8)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{2}{2\sqrt{10}}\quad =\large\boxed{\dfrac{\sqrt{10}}{10}}[/tex]

Note: 2² + 6² = hypotenuse²   →   hypotenuse = 2√10

varshamittal029

5) cosθ = [tex]\frac{\sqrt{11} }{6}[/tex] , 6) cosθ = [tex]\frac{8}{17}[/tex] , 7) tanθ = [tex]\frac{4}{3}[/tex] , 8) cosθ = [tex]\frac{1}{\sqrt{10} }[/tex]

What is pythagorean theorem ?

If, a be the perpendicular, b be the base & c be the hypotenuse of a right angle triangle.

Then, [tex]c=\sqrt{a^{2}+ b^{2} }[/tex]

What are the trigonometric functions ?

cosθ = (Side adjacent to θ)/(Hypotenuse)

sinθ = (Side opposite to θ)/(Hypotenuse)

tanθ = (Side opposite to θ)/(Side adjacent to θ)

What are the values of required trigonometric functions ?

5) Hypotenuse = [tex]\sqrt{(3\sqrt{11}) ^{2} +15^{2} }[/tex]

                            = [tex]\sqrt{99+225}[/tex]

                            = [tex]\sqrt{324}[/tex] = 18

∴ cosθ = (Side adjacent to θ)/(Hypotenuse)

           = [tex]\frac{3\sqrt{11} }{18}[/tex] =[tex]\frac{\sqrt{11} }{6}[/tex]

6) Hypotenuse = [tex]\sqrt{8^{2} +15^{2} }[/tex]

                             = [tex]\sqrt{64+225}[/tex]

                             = [tex]\sqrt{289}[/tex] = 17

∴ cosθ = (Side adjacent to θ)/(Hypotenuse)

           = [tex]\frac{8}{17}[/tex]

7) tanθ = (Side opposite to θ)/(Side adjacent to θ)

               = [tex]\frac{20}{15}[/tex] = [tex]\frac{4}{3}[/tex]

8) Hypotenuse = [tex]\sqrt{2^{2} +6^{2} }[/tex]

                              = [tex]\sqrt{4+36}[/tex]

                              = [tex]\sqrt{40}[/tex] = [tex]2\sqrt{10}[/tex]

∴ cosθ = (Side adjacent to θ)/(Hypotenuse)

            = [tex]\frac{2}{2\sqrt{10} }[/tex] = [tex]\frac{1}{\sqrt{10} }[/tex]

Learn more about trigonometric functions here :

https://brainly.com/question/1143565

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