The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Solution

Let AB be the tower of height h metres. Let C and D are two points at a distance 4 m and 9 m respectively from the base of the lower.

Let ∠BDA = ө, then ∠BCA = (90 - ө)
In right triangle BCA, we have
$tan space left parenthesis 90 space minus straight theta right parenthesis space equals space AB over BC rightwards double arrow space space space cot space straight theta space space equals space straight h over 4 space space space space space space space space.... left parenthesis straight i right parenthesis$

In right triangle BDA, we have
$tan space straight theta space equals space straight h over 9 space space space space space space space space space space space space space space space space space.... left parenthesis ii right parenthesis$
Multiplying (i) and (ii) we get
$cot space straight theta space straight x space tan space straight theta space equals space straight h over 4 straight x straight h over 9 rightwards double arrow space space 1 space equals space straight h squared over 36 rightwards double arrow space space space straight h squared equals space 36 rightwards double arrow space space space straight h space space equals space plus-or-minus 6$
Since, h = -6 is not possible
Hence, height of the tower BC is 6 m.

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Circles

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

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