Question

question 6
note: triangle may not be drawn to scale.
suppose a = 7 and b = 9.
find an exact value or give at least two decimal places:
sin(a) =
cos(a) =
tan(a) =
sec(a) =
csc(a) =
cot(a) =

Explanation:

Step1: Find the hypotenuse \(c\) using Pythagorean theorem

By Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\). Given \(a = 7\) and \(b=9\), then \(c=\sqrt{7^{2}+9^{2}}=\sqrt{49 + 81}=\sqrt{130}\)

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

\(\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{a}{c}\). Substituting \(a = 7\) and \(c=\sqrt{130}\), \(\sin(A)=\frac{7}{\sqrt{130}}=\frac{7\sqrt{130}}{130}\approx0.61\)

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

\(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{b}{c}\). Substituting \(b = 9\) and \(c=\sqrt{130}\), \(\cos(A)=\frac{9}{\sqrt{130}}=\frac{9\sqrt{130}}{130}\approx0.79\)

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

\(\tan(A)=\frac{\text{opposite}}{\text{adjacent}}=\frac{a}{b}\). Substituting \(a = 7\) and \(b = 9\), \(\tan(A)=\frac{7}{9}\approx0.78\)

Step5: Calculate \(\sec(A)\)

\(\sec(A)=\frac{1}{\cos(A)}=\frac{c}{b}\). Since \(c=\sqrt{130}\) and \(b = 9\), \(\sec(A)=\frac{\sqrt{130}}{9}\approx1.27\)

Step6: Calculate \(\csc(A)\)

\(\csc(A)=\frac{1}{\sin(A)}=\frac{c}{a}\). Since \(c=\sqrt{130}\) and \(a = 7\), \(\csc(A)=\frac{\sqrt{130}}{7}\approx1.64\)

Step7: Calculate \(\cot(A)\)

\(\cot(A)=\frac{1}{\tan(A)}=\frac{b}{a}\). Substituting \(a = 7\) and \(b = 9\), \(\cot(A)=\frac{9}{7}\approx1.29\)

Answer:

\(\sin(A)\approx0.61\), \(\cos(A)\approx0.79\), \(\tan(A)\approx0.78\), \(\sec(A)\approx1.27\), \(\csc(A)\approx1.64\), \(\cot(A)\approx1.29\)