Question

The area of an acute triangle $ABC$ is $\Delta$, the area of its pedal triangle is '$p$', where $\cos B=\frac{2p}{\Delta }$ and $\sin B=\frac{2\sqrt{3}p}{\Delta }$. The value of $8({\cos }^{2}A\cos B+{\cos }^{2}C)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$


$△DEF$ is pedal triangle of area $p$.
Area of $△ABC=\Delta$
$\cos B=\frac{2p}{\Delta },\sin B=\frac{2\sqrt{3}p}{\Delta }...\left(1\right)$

$\because {\sin }^{2}B+{\cos }^{2}B=1$
$\Rightarrow {\left(\frac{2p}{\Delta }\right)}^2+{\left(\frac{2\sqrt{3}p}{\Delta }\right)}^2=1$
$\Rightarrow \Delta =4p...\left(2\right)$

$\Delta =\frac{1}{2}ab\sin C$
$=2R^2\sin A\cdot \sin B\cdot \sin C$
$(\because a=2R\sin A,b=2R\sin B)$

Circumradius of a pedal triangle is half the circumradius of the original triangle.
Also, angles of the pedal triangle will be ${180}^o-2A,{180}^o-2b,{180}^o-2C$
Now, area of the pedal triangle will be
$\therefore p=2{\left(\frac{R}{2}\right)}^2\sin 2A\cdot \sin 2B\cdot \sin 2C$
$=\frac{R^2}{2}\sin 2A\cdot \sin 2B\cdot \sin 2C$

From eqn(2)
$\Delta =4p$
$\Rightarrow 2R^2\sin A\cdot \sin B\cdot \sin C=4\times \frac{R^2}{2}\sin 2A\cdot \sin 2B\cdot \sin 2C$
$\Rightarrow \cos A\cdot \cos B\cdot \cos C=\frac{1}{8}...\left(3\right)$

From eqn(1) and (2),
$\cos B=\frac{2p}{\Delta }=\frac{2p}{4p}=\frac{1}{2}$
$\Rightarrow \angle B=\frac{\pi }{3}...\left(4\right)$

From eqn(3),
$\cos A\cdot \cos C=\frac{1}{4}$
$\angle B=\frac{\pi }{3}\Rightarrow \angle A+\angle C={120}^{\circ }$
$\Rightarrow \cos (A+C)=-\frac{1}{2}$
$\Rightarrow \cos A\cdot \cos C-\sin A\cdot \sin C=-\frac{1}{2}$
$\Rightarrow \frac{1}{4}-\sin A\cdot \sin C=-\frac{1}{2}$
$\Rightarrow \sin A\cdot \sin C=\frac{3}{4}$

$\cos (A-C)=\cos A\cdot \cos C+\sin A\cdot \sin C$
$\Rightarrow \cos (A-C)=\frac{1}{4}+\frac{3}{4}=1$
$\Rightarrow A-C=0$
$\Rightarrow A=C$
$\because B=\frac{\pi }{3}$
$\therefore A=B=C=\frac{\pi }{3}$

$8({\cos }^{2}A\cos B+{\cos }^{2}C)=8(\frac{1}{4}\times \frac{1}{2}+\frac{1}{4})$
$=3$