A bridge is constructed on a river with two lamppoles installed on it on either side of the bridge which are exactly opposite to each other. Since one of the lamppole is made up of material which is of poor quality, one of them breaks off during heavy rain and falls on the top pf the other pole next to it making an angle of 30° with the opposite pole. After 5 minutes, due to stormy weather it completely collapses and fall on the bridge at a distance of 4m from the foot of the pole next to it. The height of the opposite lamppole is 6m. Find the original length of the broken pole before it actually broke due to stormy weather.
Solution |||
⇒ In this figure, let the broken portion of the lamppole be AA’ be of ‘x’ length.
⇒ ∴A’C = A’G = x
⇒ From the figure, total height of the broken lamppole = x + y + 6
How to proceed further?
$A'G = A'C = x$
$A'B = x \sin(30°) + 6 = {x \over 2} + 6$
$BG = x \cos(30°) - 4 = {\sqrt3 x \over 2} - 4$.
Apply Pythagorean Theorem formula on ΔA'BG
$\begin{align} x^2 &= ({x \over 2} + 6)^2 + ({\sqrt3 x \over 2}-4)^2 \cr &= ({x^2 \over 4} + 6x + 36) + ({3x^2 \over 4} - 4\sqrt3 x + 16) \cr &= x^2 + (6-4\sqrt3)x + 52 \cr\cr x &= {26 \over 2\sqrt3-3} = {26(2\sqrt3+3) \over 3} \end{align}$
$$\text{Pole height} = {3x\over2} + 6 = 13(2\sqrt3+3)+6 = 26\sqrt3+45$$