A vertical post of 126 dcm height was bent at some point above the ground and it just touched the ground making an angle of `30^(@)` with the ground. Calculate at what height was the post bent and at what distance did it meet the ground from the foot of the post.

Explanation

Let the post was AB and it was bent at C and just touched the ground at the point D making an angle of `30^(@)` with the ground.
`therefore` AC=CD and `angleBDC = 30^(@)`.
Now from the right-angled triangle BCD we get,
`sin 30^(@) = (BC)/(CD)` [ by definition ]
or, `(1)/(2) = (AB-AC)/(CD) " " or, " " (1)/(2) = (AB - CD)/(CD)`
or, 2AB - 2CD = CD or, 2AB = 3CD
or, ` 2xx 126 = 3CD " "or, " " CD = (2xx126)/(3) = 84`.
`therefore " "AC = 84, " " therefore " "BC = AB - AC = 126 - 84 = 42 `.
Again, from the right-angled triangle BDC, we get,
`tan 30^(@) = (BC)/(BD)` [ by definition ]
or, `(1)/(sqrt(3)) = (42)/(BD) [ because" " BC = 42 ]`
or, BD = `42sqrt(3)`.
Hence the post was bent at a height of 42 dcm from the ground and it met the ground at a distanece fo ` 42sqrt(3)` dcm from the foot of the post.