Approaching a tower — elevation changes from 30° to 60° in 20 m On level ground, the angle of elevation of a tower is 30°. After walking 20 m toward the tower, the angle becomes 60°. Find the tower’s height.
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A20 √3 m
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B10 √3 m
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C10 ( √3 - 1 ) m
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DNone of these
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E—
Answer
Correct Answer: 10 √3 m
Explanation
Introduction / Context:Two observations from points along the same line to a tower give two tan equations in the same unknowns (height and the initial horizontal distance). Subtracting distances resolves both quickly.
Given Data / Assumptions:
- Initial angle θ₁ = 30°, then θ₂ = 60° after moving 20 m closer.
- Let initial distance be x, tower height h.
- Level ground; vertical tower.
Concept / Approach:tan 30° = h/x and tan 60° = h/(x − 20). Solve for x, then h. Use tan 30° = 1/√3 and tan 60° = √3.
Step-by-Step Solution:
h = x/√3 (from 30°)h = √3 (x − 20) (from 60°)Equate: x/√3 = √3(x − 20) ⇒ x = 3x − 60 ⇒ 2x = 60 ⇒ x = 30h = x/√3 = 30/√3 = 10√3 mVerification / Alternative check:At 30 m, tan 30° = h/x = (10√3)/30 = 1/√3; at 10 m (after moving 20), tan 60° = (10√3)/10 = √3. Checks out.
Why Other Options Are Wrong:20√3 is double the correct height; 10(√3 − 1) and “None” do not satisfy both tan equations.
Common Pitfalls:Using 20 as the new distance (x − 20 = 20) without solving, or mixing sine/cosine for tangent relations.
Final Answer:10 √3 m