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 = 20, c = 29
sin(a) =
cos(a) =
tan(a) =
cot(a) =
sec(a) =
csc(a) =
question help: video message instructor

Explanation:

Step1: Find the length of side \(a\)

By the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). Given \(b = 20\), \(c = 29\), then \(a=\sqrt{c^{2}-b^{2}}=\sqrt{29^{2}-20^{2}}=\sqrt{(29 + 20)(29-20)}=\sqrt{49\times9}=\sqrt{441}=21\)

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

\(\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{a}{c}\), so \(\sin(A)=\frac{21}{29}\)

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

\(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{b}{c}\), so \(\cos(A)=\frac{20}{29}\)

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

\(\tan(A)=\frac{\text{opposite}}{\text{adjacent}}=\frac{a}{b}\), so \(\tan(A)=\frac{21}{20}\)

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

\(\cot(A)=\frac{\text{adjacent}}{\text{opposite}}=\frac{b}{a}\), so \(\cot(A)=\frac{20}{21}\)

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

\(\sec(A)=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{c}{b}\), so \(\sec(A)=\frac{29}{20}\)

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

\(\csc(A)=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{c}{a}\), so \(\csc(A)=\frac{29}{21}\)

Answer:

\(\sin(A)=\frac{21}{29}\), \(\cos(A)=\frac{20}{29}\), \(\tan(A)=\frac{21}{20}\), \(\cot(A)=\frac{20}{21}\), \(\sec(A)=\frac{29}{20}\), \(\csc(A)=\frac{29}{21}\)