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Orthocentre Problem

If $$x, y, z$$ are the distances of the vertices of the points $$A,B$$ and $$C$$ respectively in $$ΔABC$$ from the orthocentre, then simplify the expression:

$$\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$$

Note by Krishna Jha
3 years, 3 months ago

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We know that : $$x = 2RcosA$$ , $$y = 2RcosB$$ and $$z = 2RcosC$$

$$\frac{a}{x} + \frac{b}{y} + \frac{c}{z}$$

= $$\frac{2RsinA}{2RcosA} + \frac{2RsinB}{2RcosB} + \frac{2RsinC}{2RcosC}$$

=$$tanA + tanB + tanC$$ · 3 years, 3 months ago

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Nice attempt.. But if the options were the following then what would be the solution???:

$$(a)\frac{xyz}{abc}$$

$$(b)\frac{(xy+yz+zx)^{3}}{(abc)^{2}}$$

$$(c)\frac{abc}{xyz}$$ · 3 years, 3 months ago

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i think its option- c. since A+B+C=180. THUS tanA +tanB +tanC =tanAtanBtanC. but now clearly tanA=(a/x) tanB=(b/y) and tanC=(c/z). therefore, (a/x)+(b/y)+(c/z)=(abc/xyz). which is a pretty obvious. · 3 years, 3 months ago

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Thanks.. but could you tell me a good book/website would i find these formulae(those of properties of triangles) with their proof?? · 3 years, 3 months ago

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