Based on the following information
A man standing on a boat south of a light house observes his shadow to be 24 meters long, as measured at the sea level. On sailing 300 meters eastwards, he finds his shadow as 30 meters long, measured in a similar manner. The height of the man is 6 meters above sea level.
KL = lighthouse
BA = initial position man of man and BC = shadow
After moving 300 m east, DE = new position of man and EF = shadow
GivenĀ : AB = DE = 6 m
BC = 24 m and EF = 30 m and BE = 300 m
$$\triangle$$ LBE is right angled triangle (sea level).
To findĀ : KL = ?
SolutionĀ : In $$\triangle$$ KLF and $$\triangle$$ DEF
=> $$\angle KLF = \angle DEF = 90$$
$$\angle KFL = \angle DFE$$ (common angle)
=> $$\triangle KLF \sim \triangle DEF$$
=> $$\frac{KL}{DE} = \frac{LF}{EF}$$ -----------Eqn(I)
Similarly,Ā $$\triangle KLC \sim \triangle ABC$$
=>Ā $$\frac{KL}{AB} = \frac{LC}{BC}$$ ----------Eqn(II)
From eqn (I) and (II), and using AB = DE
=> $$\frac{LC}{BC} = \frac{LF}{EF}$$
=> $$\frac{LC}{24} = \frac{LF}{30}$$
=>Ā $$\frac{LC}{LF} = \frac{24}{30} = \frac{4}{5}$$
If, LC is 4 part $$\equiv$$ LF is 5 part
=> $$LB = 4x$$ and $$LE = 5x$$
$$\because$$ $$\triangle$$ LBE is right angled triangle
=> $$(LE)^2 - (LB)^2 = (BE)^2$$
=> $$25X^2 - 16X^2 = 90000$$
=> $$x^2 = \frac{90000}{9} = 10000$$
=> $$x = \sqrt{10000} = 100$$
=> $$LB = 400$$ and $$LE = 500$$
=> $$LC = LB + BC = 400 + 24 = 424$$
Now, using Eqn (II), we getĀ :Ā
=> $$KL = \frac{LC}{BC} \times AB$$
= $$\frac{424}{24} \times 6 = \frac{424}{4}$$
= $$106 m$$
