Heights and Distances Practice Test 1

Question : 7 of 40

Marks: +1, -0

A 6 ft.-tall man finds that the angle of elevation of the top of a 24 ft.-high pillar and the angle ofdepression its base are complementary angles. Then the distance between pillar and man is:

$2 √3$ ft
$4√ 3$ ft
$6 √3$ ft
$8 √3$ ft

Validate

Solution:

Let AB = 24 ft and CD = 6 ft be the heights of the pillar and man respectively.
Here, ∠ACE = α; ∠CBD = 90° – α;
Now from right triangle CDB,
BD = 6 cot (90° – α) = 6 tan α

∴\tan α={BD}/6

...(i)
From right triangle ACE,

\tan α =18/{EC}=18/{BD}

...(ii) (∵ EC = BD)
Now from (i) and (ii) we get,

{BD}/6=18/{BD}

⇒BD^2=18×6

∴BD=6√3\ft

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