Problem
Evaluate: \( \tan\left(\cos^{-1}\left(\frac{8}{17}\right)\right) \)
Solution
Let \( \theta = \cos^{-1}\left(\frac{8}{17}\right) \)
Then:
\[ \cos \theta = \frac{8}{17} = \frac{\text{Base}}{\text{Hypotenuse}} \]
So,
- Base = 8
- Hypotenuse = 17
Perpendicular:
\[ \sqrt{17^2 – 8^2} = \sqrt{289 – 64} = \sqrt{225} = 15 \]
Now, using:
\[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \]
\[ \tan \theta = \frac{15}{8} \]
Therefore:
\[ \tan\left(\cos^{-1}\left(\frac{8}{17}\right)\right) = \frac{15}{8} \]
Final Answer
\[ \boxed{\frac{15}{8}} \]
Explanation
Using right triangle definitions: cos = base/hypotenuse and tan = perpendicular/base.