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Difficulty: MediumPythagorean Theorem and Special Right Triangles

A maintenance worker leans a ladder against a vertical wall such that the ladder makes a 60∘60^\circ angle with the horizontal ground, reaching a height of 153 feet15\sqrt{3}\text{ feet} up the wall. If the base of the ladder is then pulled further away from the wall until the ladder makes a 45∘45^\circ angle with the horizontal ground, how many feet further from the wall is the base of the ladder?

  1. 152−1515\sqrt{2} - 15Answer
  2. B
    153−1515\sqrt{3} - 15
  3. C
    15215\sqrt{2}
  4. D
    302−3030\sqrt{2} - 30
  5. E
    30−15330 - 15\sqrt{3}

Answer

The base of the ladder is 152−1515\sqrt{2} - 15 feet further from the wall.
In the initial position, the ladder forms a 30°-60°-90° right triangle with the wall and ground. The side opposite the 60° angle (height on the wall) is 153 ft15\sqrt{3}\text{ ft}. Using the ratio 1:3:21:\sqrt{3}:2, the shorter leg (initial distance from the wall) is 15 ft15\text{ ft}, and the hypotenuse (ladder length) is 30 ft30\text{ ft}. In the second position, the ladder forms a 45°-45°-90° triangle with hypotenuse 30 ft30\text{ ft}. The leg length (new distance from the wall) is 302=152 ft\frac{30}{\sqrt{2}} = 15\sqrt{2}\text{ ft}. The additional distance the ladder base was pulled is 152−15 ft15\sqrt{2} - 15\text{ ft}.

Step-by-Step Solution

1
Determine the initial base distance and ladder length using 30°-60°-90° triangle relationships.
Initial base distance = 15 ft15\text{ ft}, ladder length = 30 ft30\text{ ft}.
In a 30°-60°-90° right triangle, the side opposite the 60° angle is x3x\sqrt{3}. Given x3=153x\sqrt{3} = 15\sqrt{3}, the shorter leg (initial base distance) is x=15 ftx = 15\text{ ft} and the hypotenuse (ladder length) is 2x=30 ft2x = 30\text{ ft}.
2
Determine the new base distance using 45°-45°-90° triangle relationships.
New base distance = 152 ft15\sqrt{2}\text{ ft}.
When the ladder (hypotenuse of 30 ft) makes a 45° angle with the ground, it forms a 45°-45°-90° right triangle where hypotenuse = leg×2\text{leg} \times \sqrt{2}. Thus, the new base distance is 302=152 ft\frac{30}{\sqrt{2}} = 15\sqrt{2}\text{ ft}.
3
Calculate how much further the base was pulled from the wall.
152−15 ft15\sqrt{2} - 15\text{ ft}.
Subtract the initial base distance (15 ft15\text{ ft}) from the new base distance (152 ft15\sqrt{2}\text{ ft}).

Key Concept

Special Right Triangle Ratios (30°-60°-90° and 45°-45°-90°)
Estimated Time:1m 30s
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Topics

Pythagorean Theorem and Special Right Triangles

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