Question

the shadow of a tower standing on a level ground found to be 40 m .longer when the sun's altitude is 30 ° then when it was 60° . find the height of the tower.

Answer 1

Answer:

let the tower ab

<acb=60'

l of shadow =bc

<adb=30'

in Δabc

tan60'=ab/bc

bc=√3bc

now in Δ abd

tan30'=ab/bc+30

1/√3=√3bc/bc+40

bc+40=3bc

40=2bc

bc=20m

ab=20√3

so the height of tower is 20√3

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