Question

refer to right triangle abc with c = 90° as shown in the figure (not drawn to scale).
use the given information to find the six trigonometric functions of a. answer exactly.
b = 5, c = 13
sin(a) =
cos(a) =
tan(a) =
cot(a) =
sec(a) =
csc(a) =

Explanation:

Step1: Find side \(a\) using Pythagorean theorem

In a right - triangle \(ABC\) with \(C = 90^{\circ}\), by the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). Given \(b = 5\) and \(c = 13\), we have \(a=\sqrt{c^{2}-b^{2}}\).

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Step2: Calculate \(\sin(A)\)

\(\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}\), for angle \(A\), the opposite side is \(a = 12\) and the hypotenuse is \(c = 13\). So \(\sin(A)=\frac{a}{c}=\frac{12}{13}\)

Step3: Calculate \(\cos(A)\)

\(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}\), for angle \(A\), the adjacent side is \(b = 5\) and the hypotenuse is \(c = 13\). So \(\cos(A)=\frac{b}{c}=\frac{5}{13}\)

Step4: Calculate \(\tan(A)\)

\(\tan(A)=\frac{\text{opposite}}{\text{adjacent}}\), for angle \(A\), the opposite side is \(a = 12\) and the adjacent side is \(b = 5\). So \(\tan(A)=\frac{a}{b}=\frac{12}{5}\)

Step5: Calculate \(\cot(A)\)

\(\cot(A)=\frac{\text{adjacent}}{\text{opposite}}\), for angle \(A\), the adjacent side is \(b = 5\) and the opposite side is \(a = 12\). So \(\cot(A)=\frac{b}{a}=\frac{5}{12}\)

Step6: Calculate \(\sec(A)\)

\(\sec(A)=\frac{\text{hypotenuse}}{\text{adjacent}}\), for angle \(A\), the hypotenuse is \(c = 13\) and the adjacent side is \(b = 5\). So \(\sec(A)=\frac{c}{b}=\frac{13}{5}\)

Step7: Calculate \(\csc(A)\)

\(\csc(A)=\frac{\text{hypotenuse}}{\text{opposite}}\), for angle \(A\), the hypotenuse is \(c = 13\) and the opposite side is \(a = 12\). So \(\csc(A)=\frac{c}{a}=\frac{13}{12}\)

Answer:

\(\sin(A)=\frac{12}{13}\), \(\cos(A)=\frac{5}{13}\), \(\tan(A)=\frac{12}{5}\), \(\cot(A)=\frac{5}{12}\), \(\sec(A)=\frac{13}{5}\), \(\csc(A)=\frac{13}{12}\)