The angle of elevation of the top of a tower from the two points PP and QQ at distances of aa and bb respectively from the base and in the same straight line with it are complementary. Prove that the height of the tower is abab where a>ba>b.


Answer:


Step by Step Explanation:
  1. Let ABAB be the tower of height hh and APBAPB be θ.θ.
    As APBAPB and AQBAQB are complementary angles, AQB=90θ.AQB=90θ.

    The image below represents the given situation.
    P B Q A h b a θ 90 - θ
  2. Now, from right-angled triangle APBAPB, we have tanθ=ABPBtanθ=hah=atanθ(i)
  3. Now, from right-angled triangle AQB, we have tan(90θ)=ABBQcotθ=hb[tan(90θ)=cotθ]h=b cotθ=btanθ(ii)
  4. On multiplying eq (i) and eq (ii), we get h2=abh=ab
  5. Therefore, the height of the tower is ab.

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