If in a triangle ABC, sin A, sin B, sin C are | vedclass

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

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the altitudes are in $H.P.$

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(B) Given that $\sin A, \sin B, \sin C$ are in $A.P.$By the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, where $R$ is the circumradius.Thus, $a, b, c$ are in $A.P.$Let $p_1, p_2, p_3$ be the altitudes corresponding to sides $a, b, c$ respectively.The area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.This implies $p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}$.Since $a, b, c$ are in $A.P.$, their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $H.P.$Therefore, $\frac{2\Delta}{a}, \frac{2\Delta}{b}, \frac{2\Delta}{c}$ are in $H.P.$Hence, the altitudes $p_1, p_2, p_3$ are in $H.P.$

If in a triangle $ABC$, $\sin A, \sin B, \sin C$ are in $A.P.$, then

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