TS EAMCET · Maths · Heights and Distances
\(P\) is a point on the segment joining the feet of two vertical poles of heights \(a\) and \(b\). The angles of elevation of the tops of the poles from \(P\) are \(45^{\circ}\) each. Then, the square of the distance between the tops of the poles is
- A \(\frac{a^2+b^2}{2}\)
- B \(a^2+b^2\)
- C \(2\left(a^2+b^2\right)\)
- D \(4\left(a^2+b^2\right)\)
Answer & Solution
Correct Answer
(C) \(2\left(a^2+b^2\right)\)
Step-by-step Solution
Detailed explanation
In \(\triangle A P D\), \(\tan 45^{\circ}=\frac{a}{A P} \Rightarrow A P=a\) and in \(\triangle B P C\), \(\tan 45^{\circ}=\frac{b}{P B} \Rightarrow P B=b\) \(\therefore D E=a+b\) and \(C E=b-a\) In \(\triangle D E C\),…
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