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tan 30° = 30/d, hence d = 30√3 m.
tan 45° = 7/d, so d = 7 m.
Distance = 20/tan 30° = 20√3 m. Tower height = 20√3 tan 60° = 60 m.
Height = 10 tan 45° = 10 m.
Height = 20 tan 30° = 20/√3 m.
Distances are 4 cot 45° = 4 m and 4 cot 30° = 4√3 m. Width = 4 + 4√3 = 4(1 + √3) m.
Width = 6 cot 30° + 6 cot 60° = 6√3 + 6/√3 = 6(√3 + 1/√3) m.
tan 30° = 50/d, so d = 50√3 m.
Initial distance = 30 m and final distance = 90 m. Therefore, distance moved = 60 m.
Let nearer distance be x. Then h = x√3 and h = (x + 40)/√3. Solving gives x = 20/√3 and h = 20√3 m.
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