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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Two towers face each other separated by a distance of 25 meters. As seen from the top of the first tower, the angle of depression of the second tower’s base is 60° and that of the top is 30°. The height (in meters) of the second tower is
Solution
Two Towers – Height of the Second Tower
Distance between towers’ bases = 25 m
Let the first tower be \(AB\) (top \(A\), base \(B\)) and the second tower be \(CD\) (top \(C\), base \(D\)). The bases \(B\) and \(D\) are 25 m apart.
From the top \(A\): angle of depression to base \(D\) is \(60^\circ\) and to top \(C\) is \(30^\circ\).
- From right \(\triangle ABD\): \[ \tan 60^\circ=\frac{AB}{BD}=\frac{AB}{25}\;\Rightarrow\; AB=25\sqrt{3}. \]
- From right \(\triangle ACD\): vertical difference \(=AB-CD\) and horizontal \(=25\). \[ \tan 30^\circ=\frac{AB-CD}{25}=\frac{1}{\sqrt{3}} \;\Rightarrow\; AB-CD=\frac{25}{\sqrt{3}}. \]
- Substitute \(AB=25\sqrt{3}\): \[ 25\sqrt{3}-CD=\frac{25}{\sqrt{3}} \;\Rightarrow\; CD=25\!\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) =\frac{50}{\sqrt{3}}\;\text{m}. \]
Answer: \( \displaystyle CD=\frac{50}{\sqrt{3}} \approx 28.87\ \text{m} \)
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