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Let A(0,0), B(b,0), C(0,c) be vertices of a triangle. If BC = a and BC is divided into n equal parts (segments) (n odd integer). Let ‘ $θ$ ’ be the angle subtended by that segment of hypotenuse BC, which contains the mid point of BC at A. If h be the length of the altitude to the hypotenuse of the triangle, and if $\tan θ = \frac{p n h}{(q n^{2} - r) a}$ where $p, q, r \in N$ and $G . C . D (p, q, r) = 1$ then which is/are correct

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$p > q$

$p < q$

$p + q = 5$

$r = 2$

answer is A, C.

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Detailed Solution

$P = (\frac{n + 1}{2} b, \frac{n - 1}{2} c)$

$Q = (\frac{n - 1}{2} b, \frac{n + 1}{2} c)$ $m_{1} = \overset{\leftrightarrow}{A P} s l o p e = \frac{c}{b} \frac{(n - 1)}{(n + 1)}$

$m_{2} = \overset{\leftrightarrow}{A Q} s l o p e = \frac{c}{b} (\frac{n + 1}{n - 1})$ $∴ \tan θ = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} = \frac{4 n b c}{(n^{2} - 1) (b^{2} + c^{2})}$

$= \frac{4 n b c}{(n^{2} - 1) a^{2}} = \frac{4 n h}{(n^{2} - 1) a} (∵ h = \frac{b c}{a})$

$∴ p = 4, q = 1, r = 1$

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