question_answer

If in a\[\Delta \,\,PQR\],\[\sin P,\,\,\sin Q,\,\,\sin R\]are in\[AP\], then

A)  the altitudes are in\[AP\]

B)  the altitudes are in\[HP\]

C)  the medians are in\[GP\]

D)  the medians are in\[AP\]

Correct Answer: B

Solution :

By sine rule                 \[\frac{a}{\sin P}=\frac{b}{\sin Q}=\frac{c}{\sin R}=k\]    (say) Also,      \[\frac{1}{2}a{{p}_{1}}=\Delta \] \[\Rightarrow \]               \[\frac{2\Delta }{a}={{p}_{1}}\] \[\Rightarrow \]               \[{{p}_{1}}=\frac{2\Delta }{k\sin P}\] Similarly,\[{{p}_{1}}-\frac{2\Delta }{k\sin Q}\]and\[{{p}_{3}}=\frac{2\Delta }{k\sin R}\] Since,\[\sin P,\,\,\sin Q,\,\,\sin R\]are in\[AP\]. \[\therefore \]  \[\frac{1}{\sin P}\cdot \frac{1}{\sin Q}\cdot \frac{1}{\sin R}\]are in\[HP\] \[\Rightarrow \]               \[\frac{2\Delta }{k\sin P},\,\,\frac{2\Delta }{k\sin Q},\,\,\frac{2\Delta }{k\sin R}\]are in\[HP\] \[\Rightarrow \,\,{{p}_{1}},\,{{p}_{2}},\,{{p}_{3}}\] are in HP.