- 20 m
- 10 m
- 15 m
- 17 m

Option 4 : 17 m

Tanθ = 8/32 = ¼

Tan2θ = 8/x = 2Tanθ/(1 – Tan^{2} θ) = 2 × ¼ /(1 – 1/16) = 8/15

Thus, 8/x = 8/15

⇒ x = 15 m

∴ Decrease in shadow = 32 – 15 = 17 m

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