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Distances between Special Points in a Triangle

Let in a ABC,...

Question

Let in a $△ A B C,$ $x, y, z$ are the lengths of perpendicular drawn from the vertices of the triangle to the opposite sides $a, b, c .$ Then the value of $\frac{c sin B + b sin C}{x} + \frac{a sin C + c sin A}{y} + \frac{b sin A + a sin B}{z}$ is equal to

A

6.00

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B

6.0

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C

6

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Solution

We know perpendicular distances :
$x = c sin B = b sin C, y = a sin C = c sin A, z = b sin A = a sin B \dots (i)$
Now,
$\frac{c sin B + b sin C}{x} + \frac{a sin C + c sin A}{y} + \frac{b sin A + a sin B}{z} = \frac{x + x}{x} + \frac{y + y}{y} + \frac{z + z}{z} [using (i)] = 2 + 2 + 2 = 6$

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Distances between Special Points in a Triangle

Standard XII Mathematics

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