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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