TS ICET 24 May 2019 Shift 1 Paper

Question : 121 of 200

Marks: +1, -0

The angles of elevation of a tower from the top and bottom of a building of height 30 meters are 30° and 60° respectively. Then the height of the tower (in meters) is:

Validate

Solution:

In ΔACE

\tan60°={CE}/{AC}

√3=h/{AC}

=h/√3

In ΔBDE,

\tan30° ={\ED}/{\BD}

\;{1}/{√{3}}=\;{h-30}/{{h}/{√{3}}}

⇒ h=3 h-90

2 h=90 ⇒ h=45 m

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