Question Number 78877 by M±th+et£s last updated on 21/Jan/20
$${if}\:\:\:\frac{{sin}\left({A}\right)}{{sin}\left({B}\right)}=\frac{{sin}\left({D}\right)}{{sin}\left({C}\right)} \\ $$$${and}\:{A}+{B}={C}+{D} \\ $$$$ \\ $$$${then}\:{prove}\:{that}\:\: \\ $$$${A}+{B}=\mathrm{180} \\ $$$${C}+{D}=\mathrm{180} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by mr W last updated on 21/Jan/20
$${you}\:{can}\:{not}\:{prove}\:{sir},\:{because}\:{the} \\ $$$${statement}\:{is}\:{not}\:{correct}.\:{in}\:{fact}\:{there} \\ $$$${are}\:{infinite}\:{possibilities}!\:\: \\ $$$${e}.{g}.\:{A}+{B}={C}+{D}={n}×\mathrm{180}° \\ $$$${e}.{g}.\:{A}+{B}={C}+{D}={any}\:{value}\:{if}\:{A}={D},\:{B}={C}. \\ $$
Commented by M±th+et£s last updated on 21/Jan/20
$${you}\:{are}\:{right}\:{sir} \\ $$
Commented by M±th+et£s last updated on 21/Jan/20
$${its}\:{my}\:{fault}\:{sorry} \\ $$
Commented by M±th+et£s last updated on 21/Jan/20
$${if}\:\frac{{sin}\left({A}\right)}{{sin}\left({B}\right)}=\frac{{sin}\left({D}\right)}{{sin}\left({C}\right)} \\ $$$${and}\:{A}+{B}={D}+{C}\:\:\:\:\:\:{A}\neq{c} \\ $$$$ \\ $$$${prove}\:{that}\: \\ $$$${A}+{B}=\mathrm{180}{n}\:\:\:\:\:{n}\in{Z} \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 21/Jan/20
$${is}\:{it}\:{right}\:{now}\:{sir} \\ $$
Commented by mr W last updated on 21/Jan/20
$${no}! \\ $$$${all}\:{these}\:{fulfill}\:{the}\:{eqn}.: \\ $$$${A}=\mathrm{5}°,\:{B}=\mathrm{6}°\:\Rightarrow{A}+{B}=\mathrm{11}° \\ $$$${C}=\mathrm{6}°,\:{D}=\mathrm{5}°\:\Rightarrow{C}+{D}=\mathrm{11}° \\ $$$${A}=\mathrm{185}°,\:{B}=\mathrm{6}°\:\Rightarrow{A}+{B}=\mathrm{191}° \\ $$$${C}=\mathrm{186}°,\:{D}=\mathrm{5}°\:\Rightarrow{C}+{D}=\mathrm{191}° \\ $$
Answered by $@ty@m123 last updated on 21/Jan/20
![Given ((sin(A))/(sin(B)))=((sin(D))/(sin(C))) ⇒ ((sinA+sin B)/(sinA−sin B))=((sin(D)+sin C)/(sinD−sin C)) ⇒ ((2sin((A+B)/2)cos ((A−B)/2))/(2cos ((A+B)/2)sin ((A−B)/2)))=((2sin((D+C)/2)cos ((D−C)/2))/(2cos ((D+C)/2)sin ((D−C)/2))) ⇒ ((cos ((A−B)/2))/(sin ((A−B)/2)))=((cos ((D−C)/2))/(sin ((D−C)/2))) {∵A+B=C+D ⇒cot (A−B)=cot (D−C) ....(i) ⇒A−B=D−C ∨A−B=π+(D−C) ⇒[A=D∧B=C]∨[A+C−(B+D)=π] I think there is some typo in question. Pl. check.](https://www.tinkutara.com/question/Q78884.png)
$${Given}\:\:\frac{{sin}\left({A}\right)}{{sin}\left({B}\right)}=\frac{{sin}\left({D}\right)}{{sin}\left({C}\right)} \\ $$$$\Rightarrow\:\frac{{sinA}+\mathrm{sin}\:{B}}{{sinA}−\mathrm{sin}\:{B}}=\frac{{sin}\left({D}\right)+\mathrm{sin}\:{C}}{{sinD}−\mathrm{sin}\:{C}} \\ $$$$\Rightarrow\:\frac{\mathrm{2}{sin}\frac{{A}+{B}}{\mathrm{2}}\mathrm{cos}\:\:\frac{{A}−{B}}{\mathrm{2}}}{\mathrm{2cos}\:\frac{{A}+{B}}{\mathrm{2}}\mathrm{sin}\:\frac{{A}−{B}}{\mathrm{2}}}=\frac{\mathrm{2}{sin}\frac{{D}+{C}}{\mathrm{2}}\mathrm{cos}\:\:\frac{{D}−{C}}{\mathrm{2}}}{\mathrm{2cos}\:\frac{{D}+{C}}{\mathrm{2}}\mathrm{sin}\:\frac{{D}−{C}}{\mathrm{2}}} \\ $$$$\Rightarrow\:\frac{\mathrm{cos}\:\:\frac{{A}−{B}}{\mathrm{2}}}{\mathrm{sin}\:\frac{{A}−{B}}{\mathrm{2}}}=\frac{\mathrm{cos}\:\:\frac{{D}−{C}}{\mathrm{2}}}{\mathrm{sin}\:\frac{{D}−{C}}{\mathrm{2}}}\:\left\{\because{A}+{B}={C}+{D}\right. \\ $$$$\Rightarrow\mathrm{cot}\:\left({A}−{B}\right)=\mathrm{cot}\:\left({D}−{C}\right)\:….\left({i}\right) \\ $$$$\Rightarrow{A}−{B}={D}−{C}\:\vee{A}−{B}=\pi+\left({D}−{C}\right) \\ $$$$\Rightarrow\left[{A}={D}\wedge{B}={C}\right]\vee\left[{A}+{C}−\left({B}+{D}\right)=\pi\right] \\ $$$${I}\:{think}\:{there}\:{is}\:{some}\:{typo}\:{in}\:{question}. \\ $$$${Pl}.\:{check}. \\ $$