Question

If $A,B,C$ and $D$ be the angles of a cyclic quadrilateral takes in order, then $\cos ({180}^o+A)+\cos ({180}^o+B)+\cos ({180}^o+D)$ is equal to

• A. $-(\cos A+\cos B+\cos D)$
• B. $2(\cos A+\cos D)$
• C. $0$
• D. $4\cos A$

Answer 1

The correct option is A $-(\cos A+\cos B+\cos D)$

$cos({180}^{\circ }+A)+cos({180}^{\circ }+B)+cos({180}^{\circ }+D)$

as A,B,C,D are angle of cyclic Qurd.

so, $\angle A+\angle C={180}^{\circ }$

$\angle B+\angle D={180}^{\circ }$

$\therefore cos({180}^{\circ }+A)=-cosA$

$cos({180}^{\circ }+B-cosB$

$cos({180}^{\circ }+D)=-cosD$

$\Rightarrow -[cosA+cosB+cosD]$