Question

For any quadrilateral $ABCD$ which of the following statement are true? $\sin (A+B)+\sin (C+D)=0$

$\cos (A+B)=\cos (C+D)$

Select the correct answer using the codes given below

• A. Only $I$
• B. Only $II$
• C. Both $I$ and $II$
• D. None of these

Answer 1

Answer: (C)

Both $I$ and $II$

In a Quadrilateral $ABCD,\{\angle A+\angle B+\angle C+\angle D={360}^{\circ }\}$

as $\angle A+\angle B={360}^{\circ }-(\angle C+\angle D)$

$\therefore sin(A+B)=sin({360}^{\circ }-(\angle C+\angle D\left)\right)$

$=-sin(\angle C+\angle D)$ ___ (1)

$\therefore sin(A+B)+sin(C+D)$

$\Rightarrow -sin(C+D)+sin(C+D)=0$

(I) is true

(II) $cos(A+B)=cos(C+D)=?$

$=cos({360}^{\circ }-(A+B\left)\right)$

$=cos(A+B)$

$\therefore cos(A+B)=cos(C+D)$

Proved.

$\therefore$ (I) & II are true

(C) is correct.