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The image shows a trigonometric problem involving a right triangle. The lengths of the legs of the triangle are \(6\) and \(9\), and we are to find the length of the hypotenuse and the tangent of the angle \( \theta \) opposite to the leg of length \(6\).

Let's start by finding the hypotenuse \( c \) using the Pythagorean theorem:

\[ c = \sqrt{6^2 + 9^2} = \sqrt{36 + 81} = \sqrt{117} = 3\sqrt{13} \]

Next, we find \(\tan(\theta)\):

\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{9} = \frac{2}{3} \]

To summarize:
1. The length of the hypotenuse \( c \) is \( 3\sqrt{13} \).
2. The tangent of the angle \( \theta \) is \( \frac{2}{3} \).

Would you like more details or have any questions?

Here are five more questions you might be interested in:
1. How do you find the sine and cosine of the angle \( \theta \)?
2. Can you explain how to solve for angles using inverse trigonometric functions?
3. How do you use the unit circle to find trigonometric values?
4. What are the identities for sum and difference of angles?
5. How can you apply the Law of Sines and the Law of Cosines in non-right triangles?

**Tip:** Always double-check your calculations in trigonometric problems to ensure accuracy, especially when dealing with square roots and fractions.

Explore a trigonometric problem involving a right triangle with legs of lengths 6 and 9. Learn how to find the length of the hypotenuse using the Pythagorean theorem and calculate the tangent of the angle opposite the leg of length 6. Suitable for Grades 10-12.

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