Height of the lamp post

It would be easier to explain if I could post images here, but I don't. I hope some one can help me.

Let:

• H = the height of lamp post
• h = 1.8 m the height of the man
• d = distance between P and Q, and Q and R
• a = distance between P and the pole = length of shadow at Q

When the man is at Q, we notice that the triangle of the tip of his shadow A, the top of his head B and Q ( $\triangle ABQ$ ) is similar to that of A, the top of the lamp post C and the bottom of the lamp post D (\triangle ACD).

Therefore,

$\dfrac{CD}{BQ} = \dfrac {AQ}{AD}$

We note that $CD = H, BQ = h , AQ = a$ and $AD = AQ + \sqrt {DP^2+PQ^2} = a + \sqrt{a^2+d^2}$ .

$\Rightarrow \dfrac {H}{h} = \dfrac { a + \sqrt{a^2+d^2}}{a} = 1 + \sqrt{1+(\frac{d}{a})^2}$

Similarly, when he is at R,

$\Rightarrow \dfrac {H}{h} = \dfrac {\frac{3}{2} a+ \sqrt{a^2+(2d)^2}}{\frac {3} {2} a} = 1 + \frac {2}{3} \sqrt{1+4(\frac {d} {a})^2}$

Equating the two equation, and let $x = \frac{d}{a}$

$1 + \sqrt{1+x^2} = 1 + \frac {2}{3} \sqrt{1+4x^2}$

$3 \sqrt{1+x^2} = 2 \sqrt{1+4x^2}$

$9 + 9 x^2 = 4 + 16 x^2 \quad \Rightarrow x^2 = \frac {5}{7}$

SInce $\dfrac {H}{h} = 1 + \sqrt{1+(\frac{d}{a})^2}$

$\Rightarrow H = h(1 + \sqrt{1+x^2}) = 1.8(1+\sqrt{1+\frac{5}{7}}) = 1.8(1+\sqrt{\frac{12}{7}} ) = \boxed {4.156}$

Let

S - length of shadow at point Q

d - distance between points P and Q, and Q and R.

The imaginary line where the points P, Q and R lies at S distance from the lamp post with P being the closest to the lamp post. So, the distance between the lamp post and point Q is $\sqrt { { S }^{ 2 }+{ d }^{ 2 } }$ . Since point R is 2d away from P, the distance between the lamp post and point R is $\sqrt { { S }^{ 2 }+{ 4d }^{ 2 } }$ By similar triangles at point Q, we get:

$\frac { H }{ \sqrt { { S }^{ 2 }+{ d }^{ 2 } } +S } =\frac { 1.8 }{ S } ; H=\frac { 1.8\left( \sqrt { { S }^{ 2 }+{ d }^{ 2 } } +S \right) }{ S }$

Then at point R:

$\frac { H }{ \sqrt { { S }^{ 2 }+{ 4d }^{ 2 } } +\frac { 3 }{ 2 } S } =\frac { 1.8 }{ \frac { 3 }{ 2 } S } ; H=\frac { 1.2\left( \sqrt { { S }^{ 2 }+{ 4d }^{ 2 } } +\frac { 3 }{ 2 } S \right) }{ S }$

By equating the equations obtained above, we get the relation:

${ d }^{ 2 }=\frac { 5 }{ 7 } { S }^{ 2 }$

Finally, by substituting the relation to any of the above equations, we obtain:

$H=\frac { 1.8\left( \sqrt { { S }^{ 2 }+\frac { 5 }{ 7 } { S }^{ 2 } } +S \right) }{ S } =\frac { 1.8S\left( \sqrt { \frac { 12 }{ 7 } } +1 \right) }{ S }=\boxed { 4.156 }$

Answer
= [ (√7 + √12) / √7 ] × 1.8
= 4.15675322 m

Let PR = 2x and the shadows' length be y (also the same as the lamp post distance to the nearest roadside at P) and 1.5y.

√( y² + x² ) / y = √( y² + (2x)² ) / 1.5y
(9/4) × ( y² + x² ) = ( y² + 4x² )
9y² + 9x² = 4 × ( y² + 4x² ) = 4y² + 16x²
5y² = 7x²
y / x = √7 / √5

If lamp post height = L, then the ratio is
L / 1.8 = [ √( y² + x² ) + y ] / y = [ √( y² + (2x)² ) + 1.5y ] / 1.5y
L / 1.8 = ( √12 + √7 ) / √7 = ( √27 + 1.5√7 ) / 1.5√7
L = 1.8 × ( 1 + √(12/7) )
= 4.15675321 m
= 13' 8" ft