AP EAMCET · Maths · Straight Lines
In H is orthocentre of \(\triangle \mathrm{ABC}\) and \(\mathrm{AH}=x ; \mathrm{BH}=y ; \mathrm{CH}=\) \(z\) then \(\frac{a b c}{x y z}=\)
- A \(1\)
- B \(\frac{a+b+c}{x+y+z}\)
- C \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\)
- D \(\frac{a b+b c+c a}{x y+y z+z x}\)
Answer & Solution
Correct Answer
(C) \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\)
Step-by-step Solution
Detailed explanation
Given \(H\) is orthocentre of \(\triangle A B C\) and \(\mathrm{AH}=x, \mathrm{BH}=y, \mathrm{CH}=z\) Since, \(\mathrm{AH}=2 \mathrm{R} \cos \mathrm{A}=x\) \(\mathrm{BH}=2 \mathrm{R} \cos \mathrm{B}=y\) \(\mathrm{CH}=2 \mathrm{R} \cos c=z\) Now,…
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