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refer to right triangle abc with c = 90° as shown in the figure (not dr…

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 = 21, c = 29
sin(a) =
cos(a) =
tan(a) =
cot(a) =
sec(a) =
csc(a) =
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Explanation:

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

By Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). Given \(b = 21\) and \(c = 29\), then \(a=\sqrt{c^{2}-b^{2}}=\sqrt{29^{2}-21^{2}}=\sqrt{(29 + 21)(29-21)}=\sqrt{50\times8}=\sqrt{400}=20\)

Step2: Recall trigonometric function definitions

  • \(\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{a}{c}\)
  • \(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{b}{c}\)
  • \(\tan(A)=\frac{\text{opposite}}{\text{adjacent}}=\frac{a}{b}\)
  • \(\cot(A)=\frac{\text{adjacent}}{\text{opposite}}=\frac{b}{a}\)
  • \(\sec(A)=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{c}{b}\)
  • \(\csc(A)=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{c}{a}\)

Step3: Calculate each function

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

Answer:

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