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Trigonometry: Lighthouse and Shadows

XAT 2011 · Quantitative Ability & Data Interpretation · Hard · Trigonometry

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Passage / data set

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.

Question 1 of 2

The height of the light house above the sea level is:

  1. A.

    90 meters

  2. B.

    94 meters

  3. C.

    96 meters

  4. D.

    100 meters

  5. E.

    106 meters

Answer

E

Explanation

Using similar triangles ΔKLF∼ΔDEF\Delta KLF \sim \Delta DEF and ΔKLC∼ΔABC\Delta KLC \sim \Delta ABC, we get LC/LF=24/30=4/5LC/LF = 24/30 = 4/5. With LB2+BE2=LE2LB^2 + BE^2 = LE^2 where BE=300BE = 300, solving 25x2−16x2=90000  ⟹  x=100  ⟹  LE=500,LB=40025x^2 - 16x^2 = 90000 \implies x = 100 \implies LE = 500, LB = 400. LC=424LC = 424. KL=(424/24)×6=106KL = (424/24) \times 6 = 106 meters.

Question 2 of 2

What is the horizontal distance of the man from the lighthouse in the second position?

  1. A.

    300 meters

  2. B.

    400 meters

  3. C.

    500 meters

  4. D.

    600 meters

  5. E.

    None of the above

Answer

C

Explanation

Horizontal distance from lighthouse to man in second position LE=500LE = 500 meters.

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