Question

problem 3. consider a right triangle with side lengths with 5, 12, 13, where the hypotenuse is length 13 and θ is the angle adjacent to the leg of length 12.

• (a) sketch the right triangle with all the sides and θ labeled.
• (b) use the triangle to find all six trig functions of θ.

Explanation:

Step1: Recall trigonometric function definitions

In a right - triangle, for an angle \(\theta\), \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), \(\csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}\), \(\sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}\), \(\cot\theta=\frac{\text{adjacent}}{\text{opposite}}\). Given hypotenuse \(c = 13\), adjacent side \(a=12\), then the opposite side \(b\) can be found using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). So \(b=\sqrt{c^{2}-a^{2}}=\sqrt{13^{2}-12^{2}}=\sqrt{169 - 144}=\sqrt{25}=5\).

Step2: Calculate trigonometric functions

• \(\sin\theta=\frac{5}{13}\) (opposite over hypotenuse)
• \(\cos\theta=\frac{12}{13}\) (adjacent over hypotenuse)
• \(\tan\theta=\frac{5}{12}\) (opposite over adjacent)
• \(\csc\theta=\frac{13}{5}\) (hypotenuse over opposite)
• \(\sec\theta=\frac{13}{12}\) (hypotenuse over adjacent)
• \(\cot\theta=\frac{12}{5}\) (adjacent over opposite)

Answer:

\(\sin\theta=\frac{5}{13}\), \(\cos\theta=\frac{12}{13}\), \(\tan\theta=\frac{5}{12}\), \(\csc\theta=\frac{13}{5}\), \(\sec\theta=\frac{13}{12}\), \(\cot\theta=\frac{12}{5}\)