SSC CGL Tier II 1 Dec 2016 Paper 1

Question : 92 of 100

Marks: +1, -0

From the top of a cliff 100 metre high, the angles of depression of the top and bottom of a tower are 45° and 60° respectively. The height of the tower is

${100}/{3}(3-√{3})$ metre
${100}/{3}(√3-1)$ metre
${100}/{3}(2√3-1)$ metre
${100}/{3}(√3-√2)$ metre

Validate

Solution:

(1)

AB = Height of cliff = 100 metre.
CD = Height of tower = h metre.
∠ADE = 45°, ∠ACB = 60°
In ΔABC,

\tan 60°= { {AB}}/{ {BC}}

⇒√{3}= {100}/{ {BC}}

⇒{BC}= {100}/{√{3}} {metre}.... (i)

In ΔADE,

\tan 45°= { {AE}}/{ {DE}}

⇒ 1= {A E}/{B C}=\ {100-h}/{B C}

⇒ B C=100-h

∴{100}/{√{3}}=100-h

⇒ h=100- {100}/{\√{3}}

= {100√{3}-100}/{√{3}}

= {100(√{3}-1)}/{√{3}}= {100 √{3}(√{3}-1)}/{3}

= {100(3-√{3})}/{3}

metre

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