In a triangle PQR,if sin P, sin Q, sin R are | vedclass

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(B) Given that $\sin P, \sin Q, \sin R$ are in $A.P.$By the Sine Rule,$\frac{a}{\sin P} = \frac{b}{\sin Q} = \frac{c}{\sin R} = 2R$,where $R$ is the c

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The altitudes are in $H.P.$

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(B) Given that $\sin P, \sin Q, \sin R$ are in $A.P.$By the Sine Rule,$\frac{a}{\sin P} = \frac{b}{\sin Q} = \frac{c}{\sin R} = 2R$,where $R$ is the circumradius.This implies $\sin P = \frac{a}{2R}, \sin Q = \frac{b}{2R}, \sin R = \frac{c}{2R}$.Since $\sin P, \sin Q, \sin R$ are in $A.P.$,it follows that $a, b, c$ are in $A.P.$Let $p_1, p_2, p_3$ be the altitudes from vertices $P, Q, R$ respectively.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,$p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}$.Since $a, b, c$ are in $A.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $H.P.$Multiplying by $2\Delta$,we get $\frac{2\Delta}{a}, \frac{2\Delta}{b}, \frac{2\Delta}{c}$ are in $H.P.$Therefore,$p_1, p_2, p_3$ are in $H.P.$

In a triangle $PQR$,if $\sin P, \sin Q, \sin R$ are in $A.P.$,then

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