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The problem describes a right triangle formed by a ladder leaning against a house. The ladder length is 15 feet (the hypotenuse), and the height of the top of the ladder from the ground is 13 feet (the opposite side). We are tasked with finding the angle of elevation \( x \) (the angle between the ladder and the ground), which corresponds to the angle in the triangle.

We can use trigonometric functions to solve this. Specifically, we will use the sine function because we know the opposite side and the hypotenuse:

\[
\sin(x) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{13}{15}
\]

To find \( x \), take the inverse sine (arcsin) of \( \frac{13}{15} \):

\[
x = \arcsin\left(\frac{13}{15}\right)
\]

Let me calculate that for you.The angle of elevation \( x \) is approximately \( 60.1^\circ \), rounded to the nearest tenth of a degree.

Would you like more details, or do you have any questions?

Here are 5 related questions to explore further:
1. What would the angle of elevation be if the ladder were 20 feet long but still reached 13 feet up the wall?
2. How would the angle of elevation change if the ladder's top was only 10 feet off the ground?
3. What trigonometric ratio would you use if you knew the base and the height of the triangle instead of the hypotenuse?
4. How could you calculate the length of the base (horizontal distance) using the given information?
5. How does increasing the height the ladder reaches affect the angle of elevation?

**Tip:** To calculate angles in right triangles, always use the correct trigonometric ratio based on the sides you know (opposite, adjacent, or hypotenuse).

A 15 ft ladder leans against the side of a house. The top of the ladder is 13 ft off the ground. Find x, the angle of elevation of the ladder. Round your answer to the nearest tenth of a degree.

Learn how to calculate the angle of elevation of a 15 ft ladder leaning against a house with the top 13 ft off the ground. This problem involves trigonometric functions such as sine and arcsin, providing a step-by-step solution. Suitable for students in Grades 9-12.

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