Heights and Distances Practice Test 1

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Question : 34 of 40

Marks: +1, -0

From the top of a hill 200 m high, the angleof depression of the top and the bottom of atower are observed to be 30° and 60°. Theheight of the tower is (in m):

$166 2/3$
$133 1/3$
$200√3$
${400√3}/3$

Solution:

From ΔABC,

{AC}/{BC}=\tan60°=√3

x=200/√3

From Δ ADE,

{AE}/{DE}=\tan30°

{200-h}/x=1/√3

{200-h}/{2 00\/√3}=1/√3

h=400/3=133 1/3

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