SSC CGL Tier II 21 Sep 2014 Paper

Question : 37 of 100

Marks: +1, -0

From the top of a tower of height 180 m the angles of depression of two objects on either sides of the tower are 30° and 45°. Then the distance between the objects are

180 $(3+√{3}) m$
180 $(3-√{3}) m$
180 $(√{3}-1) m$
180 $(√{3}+1) m$

Validate

Solution:

(4)

AD is tower and B and C are two objects,
∠ABD = 30° and ∠ACD = 45°
AD = 180 metre
From ΔABD,
tan 30°

= { {AD}}/{ {BD}}

⇒{1}/{√{3}}= {180}/{ {BD}}

⇒{BD}=180√{3}

From

\Δ{ADC}

\tan 45°= { {AD}}/{ {DC}}

⇒ 1= {180}/{ {DC}} ⇒{DC}

= 180 metre

∴{BC}= {BD}+ {DC}

=180 √{3}+180

=180(√{3}+1)

metre

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