Question

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.

Answer 1

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

`c^2=a^2+b^2`

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:

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

The length of the hypotenuse is 35.

Sine formula

`\sin (angle)=\frac{\text{opposite}}{\text{hypotenuse}}`

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

`\begin{gathered} \sin (\theta)=(21)/(35) \\ \text{ Simplifying:} \\ \sin (\theta)=(3)/(5) \end{gathered}`

Cosine formula

`\cos (angle)=\frac{\text{adjacent}}{\text{hypotenuse}}`

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

`\begin{gathered} \cos (\theta)=(28)/(35) \\ \text{ Simplifying:} \\ \cos (\theta)=(4)/(5) \end{gathered}`

Tangent formula

`\tan (angle)=(opposite)/(adjacent)`

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

`\begin{gathered} \tan (\theta)=(21)/(28) \\ \text{ Simplifying:} \\ \tan (\theta)=(3)/(4) \end{gathered}`

Secant formula

`\begin{gathered} \sec (angle)=(hypotenuse)/(adjacent) \\ \text{ }alternatively\colon \\ \sec (angle)=(1)/(\cos (angle)) \end{gathered}`

Substituting with the cosine value previously found:

`\begin{gathered} \sec (\theta)=(1)/(\cos (\theta)) \\ \sec (\theta)=(1)/((4)/(5)) \\ \sec (\theta)=(5)/(4) \end{gathered}`

Cosecant formula

`\begin{gathered} csc(angle)=(hypotenuse)/(opposite) \\ \text{ }alternatively\colon \\ \csc (angle)=(1)/(\sin (angle)) \end{gathered}`

Substituting with the sine value previously found:

`\begin{gathered} \csc (\theta)=(1)/(\sin (\theta)) \\ \csc (\theta)=(1)/((3)/(5)) \\ \csc (\theta)=(5)/(3) \end{gathered}`

Cotangent formula

`\begin{gathered} cot(angle)=(adjacent)/(opposite) \\ \text{ }alternatively\colon \\ \cot (angle)=(1)/(\tan (angle)) \end{gathered}`

Substituting with the tangent value previously found:

`\begin{gathered} \cot (\theta)=(1)/(\tan (\theta)) \\ \cot (\theta)=(1)/((3)/(4)) \\ \cot (\theta)=(4)/(3) \end{gathered}`