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Answer :
Given, Length of the vertical pole = 6 m
Shadow of the pole = 4 m
Let Height of tower = h m
Length of shadow of the tower = 28 m
In \(\triangle\) ABC and \(\triangle\) DEF,
\(\angle\) C = \(\angle\) E (angular elevation of sum)
\(\angle\) B = \(\angle\) F = 90°
\(\triangle\) ABC ~ \(\triangle\) DEF (AA similarity criterion)
AB/DF = BC/EF
(If two triangles are similar corresponding sides are proportional)
6/h = 4/28
h =( 6×28)/4
h = 6 × 7
h = 42 m
Hence, the height of the tower is 42 m.