Question

question 2 of 10
a 10 foot ladder is leaning against a building. the ladder makes a 45 degree angle with the building. how far up the building does the ladder reach?
a. ( 20sqrt{2} ) feet
b. 5 feet
c. ( 5sqrt{2} ) feet
d. ( 10sqrt{2} ) feet

Explanation:

Step1: Identify the triangle type

The ladder, building, and ground form a right triangle. The ladder is the hypotenuse (10 ft), and the angle between the ladder and the building is 45°, so it's a 45-45-90 triangle (isosceles right triangle), where the legs are equal. Let the height up the building be \( x \).

Step2: Use cosine for 45°

In a right triangle, \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \). For \( \theta = 45^\circ \), adjacent is \( x \), hypotenuse is 10. So \( \cos(45^\circ)=\frac{x}{10} \). Since \( \cos(45^\circ)=\frac{\sqrt{2}}{2} \), we have \( \frac{\sqrt{2}}{2}=\frac{x}{10} \).

Step3: Solve for \( x \)

Multiply both sides by 10: \( x = 10\times\frac{\sqrt{2}}{2}=5\sqrt{2} \).

Answer:

C. \( 5\sqrt{2} \) feet