Question

Let $\Delta$ denotes the area of the acute angled $\Delta ABC$ and ${\Delta }_P$ be the area of its pedal triangle. If $\Delta =k{\Delta }_p$, then $k$ is equal to

• A. $\frac{1}{2\cos B}$
• B. $\frac{1}{2\cos A}$
• C. $\frac{1}{2\sin A\sin B\sin C}$
• D. $\frac{1}{2\cos A\cos B\cos C}$
  

Answer 1

The correct option is D $\frac{1}{2\cos A\cos B\cos C}$


In $\Delta DEF$
$\angle D={180}^0-2A$
$\angle E={180}^0-2B$
$\angle F={180}^0-2C$
$DE=c\cos C$
$EF=a\cos A$
$DF=b\cos B$
Area of $\Delta ABC=\frac{1}{2}bc\sin A\cdots \left(i\right)$
Area of pedal triangle
${\Delta }_P=\frac{1}{2}\times DE\times FD\times \sin D$
${\Delta }_P=\frac{1}{2}\times \left(c\cos C\right)\times \left(b\cos B\right)\times \sin ({180}^0-2A)$
${\Delta }_P=\frac{1}{2}\times \left(bc\right)\left(\cos C\cos B\right)\times \left(2\sin A\cos A\right)\cdots \left(ii\right)$
Using equation $\left(i\right)$ and $\left(ii\right)$
$\frac{\Delta }{{\Delta }_P}=\frac{\frac{1}{2}\left(bc\sin A\right)}{\frac{1}{2}\times \left(bc\sin A\right)\left(2\cos A\cos C\cos B\right)}$
$\Delta =\frac{1}{\left(2\cos A\cos C\cos B\right)}\times {\Delta }_P$
$\therefore k=\frac{1}{\left(2\cos A\cos B\cos C\right)}$