In $\Delta \mathrm{PQR}, \sin \mathrm{P}, \sin \mathrm{Q}$ and $\sin \mathrm{R}$ are in A.P., then
- its altitudes are in A.P.
- its altitudes are in H.P.
- its medians are in G.P.
- its medians are in A.P.
Solution
Area of $\triangle P Q R$
$\begin{aligned} & =\frac{1}{2} \times \text { base } \times \text { height } \\ & =\frac{1}{2} \times \mathrm{p}_1 \times \mathrm{a}\end{aligned}$
Area $=\frac{1}{2} \mathrm{p}_1 \mathrm{a}$
$\begin{array}{ll}
\therefore \quad & \mathrm{p}_1=2 \times \frac{\text { Area }}{\mathrm{a}} ... (i)\\
\therefore \quad & \text { similarly, } \\
& \mathrm{p}_2=\frac{2 \times \text { Area }}{\mathrm{b}} ... (ii) \\
& \mathrm{p}_3=\frac{2 \times \text { Area }}{\mathrm{c}} ... (iii)
\end{array}$
$\therefore \quad$ By sine Rule,
$\frac{a}{\sin P}=\frac{b}{\sin Q}=\frac{c}{\sin R}$
Let $\frac{a}{\sin P}=\frac{b}{\sin Q}=\frac{c}{\sin R}=k$
$\therefore \quad \sin P=\frac{a}{k}, \sin Q=\frac{b}{k}, \sin R=\frac{c}{k}$
$\sin P, \sin Q$ and $\sin R$ are in A.P.
$\therefore \quad \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P.
$\therefore \quad \frac{1}{a}, \frac{1}{b}, \frac{1}{\mathrm{c}}$ are in H.P. ... (iv)
From equations (i), (ii), (iii) and (iv), we get $\mathrm{p}_1, \mathrm{p}_2$ and $\mathrm{p}_3$ are in H.P.Asked in: MHT CET 2023 (09 May Shift 2)