If p_1, p_2, p_3 are altitudes of a triangle | vedclass

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(A) We know that the area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$. Thus,$p_1 = \frac{2\Delta}{a}$,$p_2 =

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$\frac{1}{R}$

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(A) We know that the area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$. Thus,$p_1 = \frac{2\Delta}{a}$,$p_2 = \frac{2\Delta}{b}$,and $p_3 = \frac{2\Delta}{c}$. Substituting these into the expression: $\frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3} = \frac{a \cos A}{2\Delta} + \frac{b \cos B}{2\Delta} + \frac{c \cos C}{2\Delta}$. Using the sine rule $a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$: $= \frac{2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C}{2\Delta}$ $= \frac{R(\sin 2A + \sin 2B + \sin 2C)}{2\Delta}$. In any triangle,$\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$. Also,$\Delta = 2R^2 \sin A \sin B \sin C$,so $\sin A \sin B \sin C = \frac{\Delta}{2R^2}$. Substituting this: $= \frac{R(4 \cdot \frac{\Delta}{2R^2})}{2\Delta} = \frac{2R \Delta / R^2}{2\Delta} = \frac{1}{R}$.

If $p_1, p_2, p_3$ are altitudes of a triangle $ABC$ from the vertices $A, B, C$ respectively and if $\Delta$ is the area of the triangle,$S$ is the semi-perimeter of the triangle,then find the value of $\frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3}$.

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