A 12-foot ladder is leaning against a building. The ladder makes a 45-degree angle with the ground. How far up the building does the ladder reach?

A. 6 feet
B. [tex]$12 \sqrt{2}$[/tex] feet
C. [tex]$24 \sqrt{2}$[/tex] feet
D. [tex]$6 \sqrt{2}$[/tex] feet



Answer :

Let's solve this problem using trigonometry.

1. Identify the given information:
- Length of the ladder: [tex]\( 12 \)[/tex] feet.
- Angle between the ladder and the ground: [tex]\( 45 \)[/tex] degrees.

2. Recognize that we need to find the height [tex]\( h \)[/tex] of the ladder up against the building. This height represents the vertical leg of a right triangle formed by the ladder, the building, and the distance from the base of the ladder to the building.

3. Use the sine function:
The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side (height [tex]\( h \)[/tex]) to the hypotenuse (length of the ladder [tex]\( L \)[/tex]).
[tex]\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \][/tex]
Here, [tex]\(\theta = 45^\circ\)[/tex], the opposite side is [tex]\(h\)[/tex], and the hypotenuse is [tex]\(12\)[/tex] feet.

4. Set up the equation using the sine of [tex]\( 45^\circ \)[/tex]:
[tex]\[ \sin(45^\circ) = \frac{h}{12} \][/tex]

We know from trigonometric values that:
[tex]\[ \sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707 \][/tex]

5. Substitute [tex]\(\sin(45^\circ)\)[/tex] in the equation:
[tex]\[ \frac{\sqrt{2}}{2} = \frac{h}{12} \][/tex]

6. Solve for [tex]\( h \)[/tex]:
[tex]\[ h = 12 \times \frac{\sqrt{2}}{2} \][/tex]
Simplify:
[tex]\[ h = 12 \times 0.707 \approx 12 \times \frac{\sqrt{2}}{2} \][/tex]

Simplifying further, we get:
[tex]\[ h = 12 \times \frac{\sqrt{2}}{2} = 6 \times \sqrt{2} \][/tex]

Hence, the height that the ladder reaches up the building is:

[tex]\[ \boxed{8.48528137423857 \text{ feet}} \][/tex]

However, since this height [tex]\( 8.48528137423857 \)[/tex] feet matches the calculation [tex]\( 6 \sqrt{2} \)[/tex] feet. Therefore the correct answer matching our choices is:

D. [tex]\( 6 \sqrt{2} \)[/tex] feet