The angle of elevation of the top of a tower from the two points P and Q at distances of a and b respectively from the base and in the same straight line with it are complementary. Prove that the height of the tower is ab where a>b.


Answer:


Step by Step Explanation:
  1. Let AB be the tower of height h and APB be θ.
    As APB and AQB are complementary angles, 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 APB, 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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