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JEE Mains · Maths · STD 11 - Trigonometrical equations
Let \(AD\) and \(BC\) be two vertical poles at \(A\) and \(B\) respectively on a horizontal ground. If \(AD =8 m , BC =11 m\) and \(AB =10 m ;\) then the distance (in meters) of a point \(M\) on \(AB\) from the point \(A\) such that \(M D^{2}+M C^{2}\) is minimum is
- A \(5\)
- B \(10\)
- C \(15\)
- D \(20\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
\(( MD )^{2}+( MC )^{2}= h ^{2}+64+( h -10)^{2}+121\) \(=2 h^{2}-20 h+64+100+121\) \(=2\left( h ^{2}-10 h \right)+285\) \(=2(h-5)^{2}+235\) it is minimum if \(h =5\)
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