In H is orthocentre of $\triangle \mathrm{ABC}$ and $\mathrm{AH}=x ; \mathrm{BH}=y ; \mathrm{CH}=$ $z$ then…

1. $1$
2. $\frac{a+b+c}{x+y+z}$
3. $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$
4. $\frac{a b+b c+c a}{x y+y z+z x}$

Solution

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, $\frac{a b c}{x y z}=\frac{a b c}{2 \mathrm{R} \cos \mathrm{A} \cdot 2 \mathrm{R} \cos \mathrm{B} \cdot 2 \mathrm{R} \cos \mathrm{C}}$ $\begin{aligned} & =\frac{a}{2 R} \cdot \frac{b}{2 R} \cdot \frac{c}{2 R} \cdot \frac{1}{\cos A \cdot \cos B \cdot \cos C} \\ & =\frac{\sin A \cdot \sin B \cdot \sin C}{\cos A \cdot \cos B \cdot \cos C}=\tan A \cdot \tan B \cdot \tan C\end{aligned}$ $\begin{aligned} & =\tan A+\tan B+\tan C \\ & =\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}\end{aligned}$ $=\frac{a}{2 \mathrm{R} \cos \mathrm{A}}+\frac{b}{2 \mathrm{R} \cos \mathrm{B}}+\frac{c}{2 \mathrm{R} \cos \mathrm{C}}=\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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