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Write the six trigonometric functions for a right triangle with legs of length 21 and 28, o is adjacent to the leg of length 28.

Sagot :

First, we need to find the length of the hypotenuse. The Pythagorean theorem states:

[tex]c^2=a^2+b^2[/tex]

where a and b are the legs and c is the hypotenuse of a right triangle.

Substituting with a = 21 and b = 28, we get:

[tex]\begin{gathered} c^2=21^2+28^2 \\ c^2=441+784 \\ c^2=1225 \\ c=\sqrt[]{1225} \\ c=35 \end{gathered}[/tex]

The length of the hypotenuse is 35.

Sine formula

[tex]\sin (angle)=\frac{\text{opposite}}{\text{hypotenuse}}[/tex]

Considering angle θ, the opposite leg is 21. Therefore:

[tex]\begin{gathered} \sin (\theta)=\frac{21}{35} \\ \text{ Simplifying:} \\ \sin (\theta)=\frac{3}{5} \end{gathered}[/tex]

Cosine formula

[tex]\cos (angle)=\frac{\text{adjacent}}{\text{hypotenuse}}[/tex]

Considering angle θ, the adjacent leg is 28. Therefore:

[tex]\begin{gathered} \cos (\theta)=\frac{28}{35} \\ \text{ Simplifying:} \\ \cos (\theta)=\frac{4}{5} \end{gathered}[/tex]

Tangent formula

[tex]\tan (angle)=\frac{opposite}{adjacent}[/tex]

Considering angle θ, the adjacent leg is 28 and the opposite leg is 21. Therefore:

[tex]\begin{gathered} \tan (\theta)=\frac{21}{28} \\ \text{ Simplifying:} \\ \tan (\theta)=\frac{3}{4} \end{gathered}[/tex]

Secant formula

[tex]\begin{gathered} \sec (angle)=\frac{hypotenuse}{adjacent} \\ \text{ }alternatively\colon \\ \sec (angle)=\frac{1}{\cos (angle)} \end{gathered}[/tex]

Substituting with the cosine value previously found:

[tex]\begin{gathered} \sec (\theta)=\frac{1}{\cos (\theta)} \\ \sec (\theta)=\frac{1}{\frac{4}{5}} \\ \sec (\theta)=\frac{5}{4} \end{gathered}[/tex]

Cosecant formula

[tex]\begin{gathered} csc(angle)=\frac{hypotenuse}{opposite} \\ \text{ }alternatively\colon \\ \csc (angle)=\frac{1}{\sin (angle)} \end{gathered}[/tex]

Substituting with the sine value previously found:

[tex]\begin{gathered} \csc (\theta)=\frac{1}{\sin (\theta)} \\ \csc (\theta)=\frac{1}{\frac{3}{5}} \\ \csc (\theta)=\frac{5}{3} \end{gathered}[/tex]

Cotangent formula

[tex]\begin{gathered} cot(angle)=\frac{adjacent}{opposite} \\ \text{ }alternatively\colon \\ \cot (angle)=\frac{1}{\tan (angle)} \end{gathered}[/tex]

Substituting with the tangent value previously found:

[tex]\begin{gathered} \cot (\theta)=\frac{1}{\tan (\theta)} \\ \cot (\theta)=\frac{1}{\frac{3}{4}} \\ \cot (\theta)=\frac{4}{3} \end{gathered}[/tex]