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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.
What is the horizontal distance of the man from the lighthouse in the second position?
KL = lighthouse
BA = initial position man of man and BC = shadow
After moving 300 m east, DE = new position of man and EF = shadow
Given : AB = DE = 6 m
BC = 24 m and EF = 30 m and BE = 300 m
$$\triangle$$ LBE is right angled triangle (sea level).
To find : LE = ?
Solution : 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, $$\triangle KLC \sim \triangle ABC$$
=> $$\frac{KL}{AB} = \frac{LC}{BC}$$ ----------Eqn(II)
From eqn (I) and (II), and using AB = DE
=> $$\frac{LC}{BC} = \frac{LF}{EF}$$
=> $$\frac{LC}{24} = \frac{LF}{30}$$
=> $$\frac{LC}{LF} = \frac{24}{30} = \frac{4}{5}$$
If, LC is 4 part $$\equiv$$ LF is 5 part
=> $$LB = 4x$$ and $$LE = 5x$$
$$\because$$ $$\triangle$$ LBE is right angled triangle
=> $$(LE)^2 - (LB)^2 = (BE)^2$$
=> $$25^2 - 16X^2 = 90000$$
=> $$x^2 = \frac{90000}{9} = 10000$$
=> $$x = \sqrt{10000} = 100$$
$$\therefore LE = 5 \times 100 = 500 m$$
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