Question

AL, BM and CN are diameter of the circumcircle of a triangle ABC, ${\Delta }_1,{\Delta }_2,{\Delta }_3$ and $\Delta$ are the areas of the triangles BLC, CMA, ANB and ABC respectively.

${\Delta }_1$ is equal to

• A. $2R^2$ sin A cos B cos C
• B. $2R^2$ sin A sin B cos C
• C. $2R^2$ cos A cos B sin C
• D. $2R^2$ sin A sin B sin C

Answer 1

The correct option is A $2R^2$ sin A cos B cos C

$ACL=ABL={90}^o$ (angles in a semicircle)
$ALC=ABC=B,ALB=ACB=C$
Circumcircle of triangle BLC is same as that of ABC.
$\therefore {\Delta }_1=$ Area of triangle BLC
$=\frac{BC\times BL\times CL}{4R}=\frac{a.c\cot C.b\cot B}{4R}$
$=\frac{(2R{)}^2.\sin A\sin B\sin C\cos B\cos C}{4R\sin B\sin C}$
$=2R^2\sin A\cos B\cos C$