Question

suppose abc is a right - triangle with sides of lengths a, b, and c. (see figure 3.) find the unknown side length using the pythagorean theorem and then find the values of the six trigonometric functions for angles a and b, and then find the denominators when applicable.

1. a = 5, b = 12
2. a = 6, c = 7
3. a = 3, b = 5
4. b = 7, c = 12

Explanation:

Step1: Recall Pythagorean theorem

In a right - triangle, \(c^{2}=a^{2}+b^{2}\), where \(c\) is the hypotenuse (the longest side) and \(a\) and \(b\) are the other two sides.

Step2: For problem 11

Given \(a = 5\) and \(b=12\), find \(c\).
\[c=\sqrt{a^{2}+b^{2}}=\sqrt{5^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13\]
The six trigonometric functions for an angle \(\theta\) (opposite side \(a = 5\), adjacent side \(b = 12\), hypotenuse \(c = 13\)):
\(\sin\theta=\frac{a}{c}=\frac{5}{13}\), \(\cos\theta=\frac{b}{c}=\frac{12}{13}\), \(\tan\theta=\frac{a}{b}=\frac{5}{12}\), \(\csc\theta=\frac{c}{a}=\frac{13}{5}\), \(\sec\theta=\frac{c}{b}=\frac{13}{12}\), \(\cot\theta=\frac{b}{a}=\frac{12}{5}\)

Step3: For problem 12

Given \(a = 3\) and \(b = 5\), find \(c\).
\[c=\sqrt{a^{2}+b^{2}}=\sqrt{3^{2}+5^{2}}=\sqrt{9 + 25}=\sqrt{34}\]
The six trigonometric functions for an angle \(\theta\) (opposite side \(a = 3\), adjacent side \(b = 5\), hypotenuse \(c=\sqrt{34}\)):
\(\sin\theta=\frac{a}{c}=\frac{3}{\sqrt{34}}=\frac{3\sqrt{34}}{34}\), \(\cos\theta=\frac{b}{c}=\frac{5}{\sqrt{34}}=\frac{5\sqrt{34}}{34}\), \(\tan\theta=\frac{a}{b}=\frac{3}{5}\), \(\csc\theta=\frac{c}{a}=\frac{\sqrt{34}}{3}\), \(\sec\theta=\frac{c}{b}=\frac{\sqrt{34}}{5}\), \(\cot\theta=\frac{b}{a}=\frac{5}{3}\)

Step4: For problem 13

Given \(a = 6\) and \(c = 7\), find \(b\).
\[b=\sqrt{c^{2}-a^{2}}=\sqrt{7^{2}-6^{2}}=\sqrt{49 - 36}=\sqrt{13}\]
The six trigonometric functions for an angle \(\theta\) (opposite side \(a = 6\), adjacent side \(b=\sqrt{13}\), hypotenuse \(c = 7\)):
\(\sin\theta=\frac{a}{c}=\frac{6}{7}\), \(\cos\theta=\frac{b}{c}=\frac{\sqrt{13}}{7}\), \(\tan\theta=\frac{a}{b}=\frac{6}{\sqrt{13}}=\frac{6\sqrt{13}}{13}\), \(\csc\theta=\frac{c}{a}=\frac{7}{6}\), \(\sec\theta=\frac{c}{b}=\frac{7}{\sqrt{13}}=\frac{7\sqrt{13}}{13}\), \(\cot\theta=\frac{b}{a}=\frac{\sqrt{13}}{6}\)

Step5: For problem 14

Given \(b = 7\) and \(c = 12\), find \(a\).
\[a=\sqrt{c^{2}-b^{2}}=\sqrt{12^{2}-7^{2}}=\sqrt{144 - 49}=\sqrt{95}\]
The six trigonometric functions for an angle \(\theta\) (opposite side \(a=\sqrt{95}\), adjacent side \(b = 7\), hypotenuse \(c = 12\)):
\(\sin\theta=\frac{a}{c}=\frac{\sqrt{95}}{12}\), \(\cos\theta=\frac{b}{c}=\frac{7}{12}\), \(\tan\theta=\frac{a}{b}=\frac{\sqrt{95}}{7}\), \(\csc\theta=\frac{c}{a}=\frac{12}{\sqrt{95}}=\frac{12\sqrt{95}}{95}\), \(\sec\theta=\frac{c}{b}=\frac{12}{7}\), \(\cot\theta=\frac{b}{a}=\frac{7}{\sqrt{95}}=\frac{7\sqrt{95}}{95}\)

Answer:

For problem 11: \(c = 13\), trigonometric functions as above.
For problem 12: \(c=\sqrt{34}\), trigonometric functions as above.
For problem 13: \(b=\sqrt{13}\), trigonometric functions as above.
For problem 14: \(a=\sqrt{95}\), trigonometric functions as above.