Question

In a $△ABC,$ the line segment $AD,BE$ and $CF$ are three altitudes. If $R$ is the circumradius of the $△ABC,$ then a side of the $△DEF$ will be

• A. $R\sin 2A$
• B. $\cos B$
• C. $a\sin A$
• D. $b\cos B$

Answer 1

Answer: (A), (D)

$R\sin 2A$
$b\cos B$

From geometry $\angle AOF=B,\angle AOE=C$

Also, $OF=b\cos A.\tan ({90}^0-B)=b\cos A.\cot B=2R\cos A.\cos B$

Similarly, $OE=2R\cos A\cos C$

In $△OEF,\cos (B+C)=\frac{{OE}^2+{OF}^2-{EF}^2}{2.OE.OF}$

$\Rightarrow -\cos A=\frac{4R^2{\cos }^{2}A({\cos }^{2}B+{\cos }^{2}C)-{EF}^2}{8R^2{\cos }^{2}A\cos B\cos C}$

$\Rightarrow {EF}^2=4R^2{\cos }^{2}A[{\cos }^{2}B+{\cos }^{2}C+2\cos A\cos B\cos C]$

$=R^2{\cos }^{2}A{\sin }^{2}A$

$($ because in $△ABC,{\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=1-2\cos A\cos B\cos C)$

$\therefore EF=R\sin 2A=\frac{a}{2\sin A}\sin 2A=a\cos A$

Similarly, $DF=b\cos B$