Question

An archer, from the point A, is aiming at a target, placed on the top of a 56 feet tall tree. The base of the tree is at the point C. From the archer's position, the angle of elevation to the target is 45° from his eye level. The archer, facing the tree, moves backwards on the straight line joining the points A and C, to a new position at the point B. From the point B, the angle of elevation from his eye level to the target becomes 30°.
How far did the archer move from A to B (in feet) if his eye level is at a height of 6 feet from the ground?

• A. \( \frac{50}{\sqrt{3}} \)
• B. \( 50 \sqrt{3} \)
• C. \( 56 \sqrt{3} \)
• D. \( 50(\sqrt{3} - 1) \)
• E. \( 50(1 - \frac{1}{\sqrt{3}}) \)

Hint:

In problems involving angles of elevation, the tangent function can help you relate the height and distance from the object being viewed.

Answer 1

The Correct Option is A

Step 1: Visualize the problem.
The tree height is 56 feet, and the archer's eye level is 6 feet above the ground. So, the effective height of the tree from the archer's eye level is: \[ \text{Height} = 56 - 6 = 50 \, \text{feet}. \] The angle of elevation at point A is 45°, and at point B is 30°.
Step 2: Use the tangent function for point A.
For point A, the angle of elevation is 45°. Using the tangent function: \[ \tan 45^\circ = \frac{\text{Height of tree}}{\text{Distance from point A to the tree}} = 1. \] Thus, the distance from point A to the base of the tree (denoted as \( x_1 \)) is: \[ x_1 = 50 \, \text{feet}. \]
Step 3: Use the tangent function for point B.
For point B, the angle of elevation is 30°. Using the tangent function: \[ \tan 30^\circ = \frac{\text{Height of tree}}{\text{Distance from point B to the tree}} = \frac{1}{\sqrt{3}}. \] Thus, the distance from point B to the base of the tree (denoted as \( x_2 \)) is: \[ x_2 = 50 \sqrt{3} \, \text{feet}. \]
Step 4: Calculate the distance moved.
The distance the archer moves from point A to point B is the difference in distances from the tree: \[ \text{Distance moved} = x_2 - x_1 = 50 \sqrt{3} - 50 = 50 (\sqrt{3} - 1). \]
Step 5: Conclusion.
The correct answer is \( \boxed{\frac{50}{\sqrt{3}}} \). Thus, the archer moved \( \frac{50}{\sqrt{3}} \) feet.