Question

A man standing south of a lamp post, observes his shadow on the horizontal plane to be 24 ft. long. On walking 300 ft. eastwards, he finds his shadow to be 30 ft long. If his height is 6 ft, find the height of the lamp post in ft.

Answer 1

Let, $OP$ be the lamp post of height $h$, $AB$ be the first position of the man with shadow,

$AC=24\text{ft}$ and A'B' be the second position of the man with shadow $A^{\prime }{C}^{\prime }=30\text{ft}$
Then $AB=A^{\prime }{B}^{\prime }=6\text{ft}$ and $A{A}^{\prime }=300\text{ft}$
Let, $OA=x$
Now, $△POC$ and $△BAC$ are similar.

The ratio of corresponding sides of the similar triangles are equal.
$⟹\frac{OP}{AB}=\frac{OC}{AC}$

$\frac{h}{6}=\frac{x+24}{24}$

On cross multiplication:

$4h-24=x$

Similarly, $△PO{C}^{\prime }$ and $△B^{\prime }{A}^{\prime }{C}^{\prime }$ are similar:

The ratio of corresponding sides of the similar triangles are equal.
$⟹\frac{OP}{A^{\prime }{B}^{\prime }}=\frac{O{C}^{\prime }}{A^{\prime }{C}^{\prime }}$

$\frac{h}{6}=\frac{O{A}^{\prime }+30}{30}$

$5h=\sqrt{x^2+{300}^2}+30$

Now, from right angled triangle $△OA{A}^{\prime }$:
${(5h-30)}^2={(4h-24)}^2+{300}^{2}9h^2-108h-89676=0h^2-12h-9964=0$

$h=\frac{12\pm \sqrt{144+39856}}{2}$

$=6\pm 100$

$\Rightarrow h=106\text{ft}$