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Question Number 78877 by M±th+et£s last updated on 21/Jan/20

$$if\frac{sin\left(A\right)}{sin\left(B\right)}=\frac{sin\left(D\right)}{sin\left(C\right)}$$$$andA+B=C+D$$$$$$$$thenprovethat$$$$A+B=180$$$$C+D=180$$$$$$

Commented by mr W last updated on 21/Jan/20

$$youcannotprovesir,becausethe$$$$statementisnotcorrect.infactthere$$$$areinfinitepossibilities!$$$$e.g.A+B=C+D=n\times 180°$$$$e.g.A+B=C+D=anyvalueifA=D,B=C.$$

Commented by M±th+et£s last updated on 21/Jan/20

$$youarerightsir$$

Commented by M±th+et£s last updated on 21/Jan/20

$$itsmyfaultsorry$$

Commented by M±th+et£s last updated on 21/Jan/20

$$if\frac{sin\left(A\right)}{sin\left(B\right)}=\frac{sin\left(D\right)}{sin\left(C\right)}$$$$andA+B=D+CA\ne c$$$$$$$$provethat$$$$A+B=180nn\in Z$$$$$$

Commented by M±th+et£s last updated on 21/Jan/20

$$isitrightnowsir$$

Commented by mr W last updated on 21/Jan/20

$$no!$$$$allthesefulfilltheeqn.:$$$$A=5°,B=6°\Rightarrow A+B=11°$$$$C=6°,D=5°\Rightarrow C+D=11°$$$$A=185°,B=6°\Rightarrow A+B=191°$$$$C=186°,D=5°\Rightarrow C+D=191°$$

Answered by $@ty@m123 last updated on 21/Jan/20

$$Given\frac{sin\left(A\right)}{sin\left(B\right)}=\frac{sin\left(D\right)}{sin\left(C\right)}$$$$\Rightarrow \frac{sinA+\sin B}{sinA-\sin B}=\frac{sin\left(D\right)+\sin C}{sinD-\sin C}$$$$\Rightarrow \frac{2sin\frac{A+B}{2}\cos \frac{A-B}{2}}{2\cos \frac{A+B}{2}\sin \frac{A-B}{2}}=\frac{2sin\frac{D+C}{2}\cos \frac{D-C}{2}}{2\cos \frac{D+C}{2}\sin \frac{D-C}{2}}$$$$\Rightarrow \frac{\cos \frac{A-B}{2}}{\sin \frac{A-B}{2}}=\frac{\cos \frac{D-C}{2}}{\sin \frac{D-C}{2}}\{\because A+B=C+D$$$$\Rightarrow \cot (A-B)=\cot (D-C)....\left(i\right)$$$$\Rightarrow A-B=D-C\vee A-B=\pi +(D-C)$$$$\Rightarrow [A=D\wedge B=C]\vee [A+C-(B+D)=\pi ]$$$$Ithinkthereissometypoinquestion.$$$$Pl.check.$$

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