Let P_1, P_2, P_3 be the altitudes of a trian | vedclass

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(A) The area of the triangle is given by $\Delta = \frac{1}{2} a P_1 = \frac{1}{2} b P_2 = \frac{1}{2} c P_3$. Thus, $P_1 = \frac{2\Delta}{a}, P_2 = \

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$\frac{1}{R}$

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(A) The area of the triangle is given by $\Delta = \frac{1}{2} a P_1 = \frac{1}{2} b P_2 = \frac{1}{2} c P_3$. Thus, $P_1 = \frac{2\Delta}{a}, P_2 = \frac{2\Delta}{b}, P_3 = \frac{2\Delta}{c}$. 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 $a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$: $= \frac{2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C}{2\Delta} = \frac{R(\sin 2A + \sin 2B + \sin 2C)}{2\Delta}$. Using the identity $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$: $= \frac{R(4 \sin A \sin B \sin C)}{2\Delta}$. Since $\Delta = 2R^2 \sin A \sin B \sin C$, we have $\sin A \sin B \sin C = \frac{\Delta}{2R^2}$. Substituting this: $= \frac{4R \Delta}{2\Delta (2R^2)} = \frac{4R}{4R^2} = \frac{1}{R}$.

Let $P_1, P_2, P_3$ be the altitudes of a triangle $ABC$ from the vertices $A, B, C$ respectively. If $\Delta$ denotes the area of the triangle and $s$ is the semi-perimeter of the triangle, then $\frac{\cos A}{P_1} + \frac{\cos B}{P_2} + \frac{\cos C}{P_3} = $

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