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

Sagot :

To determine how far up the building the ladder reaches, we need to find the height where the ladder touches the building. The problem provides the following information:

- The length of the ladder is 12 feet.
- The angle between the ladder and the ground (hence with the building) is 45 degrees.

To solve this, we can use trigonometry, specifically the sine function, because we know the hypotenuse (the length of the ladder) and are solving for the opposite side (the height up the building).

Here are the steps to find the height:

1. Identify the trigonometric function:
We will use the sine function, which relates the angle to the opposite side and the hypotenuse in a right triangle.
[tex]\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \][/tex]

2. Convert the angle to radians:
Trigonometric functions in the formulas should use angles measured in radians. To convert degrees to radians:
[tex]\[ \text{Radians} = \text{Degrees} \times \left( \frac{\pi}{180} \right) \][/tex]
[tex]\[ 45 \text{ degrees} = 45 \times \left( \frac{\pi}{180} \right) = 0.7854 \text{ radians} \][/tex]

3. Set up the equation:
[tex]\[ \sin(0.7854) = \frac{\text{height}}{12} \][/tex]

4. Solve for the height:
[tex]\[ \text{height} = 12 \times \sin(0.7854) \][/tex]
We now find the value of [tex]\(\sin(0.7854)\)[/tex], which is approximately [tex]\(\sin(45^\circ) = \frac{\sqrt{2}}{2}\)[/tex].

Therefore:
[tex]\[ \text{height} = 12 \times \frac{\sqrt{2}}{2} = 12 \times 0.7071 = 8.4853 \text{ feet} \][/tex]

Thus, the height at which the ladder touches the building is approximately 8.4853 feet.

Comparing with the options provided, the closest match based on the trigonometric calculation is:
A. [tex]\(6 \sqrt{2}\)[/tex] feet since [tex]\(6 \sqrt{2} \approx 8.4853\)[/tex].

So, the answer is:
[tex]\[ \boxed{A. \; 6 \sqrt{2} \text{ feet}} \][/tex]