NIMCET 2015 Question Paper with solutions

Question : 39 of 120

Marks: +1, -0

Two towers face each other separated by a distance of25 meters. As seen from the top of the first tower, theangle of depression of the second tower’s base is 60°and that of the top is 30°. The height (in meters) of thesecond tower is;

${50}/√3$
${25}/√3$
50
25 $√3$

Validate

Solution:

Let the height of first tower is PQ.
∴ PQ = AD
In given figure

In ΔAPB,

\tan60°={AB}/{PA}

⇒√3PA=AB

...(i)
In ΔBQD,

\tan30°={BD}/{QD}

1/√3={BD}/{QD}

QD=√3BD

⇒PA=√3BD

...(ii) [∵QD=PA]

⇒PA=√3[AB-AD]

From Eq. (i) and (ii), we get

PA=√3(√3PA-AD)

PA=3PA-√3AD

√3AD=2PA

AD={2×25}/√3

[∵PA=25]

AD=50/√3

Hence, height of first tower is

50/√3m

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