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Answer :

Let BC be the pedestal of height h metres and CD be the statue of height 1.6 m. Let A be a point on the ground such that \( ∠ \ CAB \ = \ 45º \) and\( ∠ \ DAB \ = \ 60º \)

In \( ∆ \ ABC \) and \( ∆ \ ABD \), we have

\( \frac{AB}{BC} \ = \ cot45º \ \)

\( => \ \frac{AB}{h} \ = \ 1 \)

\( => \ AB \ = \ h \) (1)

and, \( \frac{BD}{AB} \ = \ tan60º \ \)

\( => \ \frac{BC + CD}{AB} \ = \ \sqrt{3} \)

\( => \ \frac{h+1.6}{h} \ = \ \sqrt{3} \ \)

\( => \ h+1.6 \ = \ h \sqrt{3} \)

\(=> h( \sqrt{3} \ - \ 1) \ = \ 1.6 \ \)

\(=> \ h \ = \ \frac{1.6}{ \sqrt{3} \ - \ 1} \ \)

\( = \ \frac{1.6}{ \sqrt{3} \ - \ 1} × \frac{ \sqrt{3} \ + \ 1}{ \sqrt{3} \ + \ 1} \)

\( = \ \frac{1.6( \sqrt{3} \ + \ 1)}{2} \ \)

\( = \ 0.8( \sqrt{3}+1) \) \( = \ 2.1856 \)

\(\therefore \) the height of the pedestal is \( 2.1856 \) m

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