QUESTION IMAGE
Question
refer to right triangle abc with as shown in the figure (not drawn to scale).
use the given information to find the six trigonometric functions of a. answer exactly.
a = 3, b = 2
sin(a) =
cos(a) =
tan(a) =
cot(a) =
sec(a) =
csc(a) =
question help: video ebook written example message instructor
Step1: Find the hypotenuse \(c\)
By the Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\). Given \(a = 3\) and \(b=2\), then \(c=\sqrt{3^{2}+2^{2}}=\sqrt{9 + 4}=\sqrt{13}\).
Step2: Calculate \(\sin(A)\)
\(\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{a}{c}\). Substituting \(a = 3\) and \(c=\sqrt{13}\), we get \(\sin(A)=\frac{3}{\sqrt{13}}=\frac{3\sqrt{13}}{13}\).
Step3: Calculate \(\cos(A)\)
\(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{b}{c}\). Substituting \(b = 2\) and \(c=\sqrt{13}\), we get \(\cos(A)=\frac{2}{\sqrt{13}}=\frac{2\sqrt{13}}{13}\).
Step4: Calculate \(\tan(A)\)
\(\tan(A)=\frac{\text{opposite}}{\text{adjacent}}=\frac{a}{b}\). Substituting \(a = 3\) and \(b = 2\), we get \(\tan(A)=\frac{3}{2}\).
Step5: Calculate \(\cot(A)\)
\(\cot(A)=\frac{\text{adjacent}}{\text{opposite}}=\frac{b}{a}\). Substituting \(a = 3\) and \(b = 2\), we get \(\cot(A)=\frac{2}{3}\).
Step6: Calculate \(\sec(A)\)
\(\sec(A)=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{c}{b}\). Substituting \(b = 2\) and \(c=\sqrt{13}\), we get \(\sec(A)=\frac{\sqrt{13}}{2}\).
Step7: Calculate \(\csc(A)\)
\(\csc(A)=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{c}{a}\). Substituting \(a = 3\) and \(c=\sqrt{13}\), we get \(\csc(A)=\frac{\sqrt{13}}{3}\).
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\(\sin(A)=\frac{3\sqrt{13}}{13}\), \(\cos(A)=\frac{2\sqrt{13}}{13}\), \(\tan(A)=\frac{3}{2}\), \(\cot(A)=\frac{2}{3}\), \(\sec(A)=\frac{\sqrt{13}}{2}\), \(\csc(A)=\frac{\sqrt{13}}{3}\)