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Option 2 : 25 (√3 + 1)
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⇒ In ΔCDB, ∠CDB = 45°
⇒ Now since, tan 45° = CB/DB = 1
⇒ CB = DB = x
⇒ Now, tan 30° = CB/AB
⇒ 1/√ 3 = x/(50 + x)
⇒ (50 + x) = √3 × x
⇒ 50 = x (√3 – 1)
⇒ x = 50 / (√3 – 1)
⇒ x = 50/(√3 – 1) × (√3 + 1) /(√3 + 1)
⇒ x = 25 (√3 + 1)
∴ Answer is 25 (√3 + 1)
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