Question

There exist a triangle $ABC$ satisfying

• A. $\tan A+\tan B+\tan C=0$
• B. $\frac{\sin A}{2}=\frac{\sin B}{3}=\frac{\sin C}{7}$
• C. ${(a+b)}^2=c^2+ab$ and $\sqrt{2}(\sin A+\cos A)=\sqrt{3}$
• D. $\sin A+\sin B=\left(\frac{\sqrt{3}+1}{2}\right),\cos A\cos B=\frac{\sqrt{3}}{4}=\sin A\sin B$

Answer 1

Answer: (C), (D)

${(a+b)}^2=c^2+ab$ and $\sqrt{2}(\sin A+\cos A)=\sqrt{3}$
$\sin A+\sin B=\left(\frac{\sqrt{3}+1}{2}\right),\cos A\cos B=\frac{\sqrt{3}}{4}=\sin A\sin B$

Option$\left(a\right)$ SInce
$\tan A+\tan B+\tan C=\tan A\tan B\tan C$
But here, $\tan A+\tan B+\tan C=0,$ is impossible.
Option$\left(b\right)$
$\frac{\sin A}{2}=\frac{\sin B}{3}=\frac{\sin C}{7}$
or $\frac{a}{2}=\frac{b}{3}=\frac{c}{7}$
$\Rightarrow \frac{a+b}{5}=\frac{c}{7}$
$\Rightarrow \frac{a+b}{c}=\frac{5}{7}<1$
$\therefore a+b<c$ is impossible.
Option$\left(c\right)$
${(a+b)}^2=c^2+ab$
$\Rightarrow a^2+b^2+2ab=c^2+ab$
$\Rightarrow a^2+b^2-c^2=-ab$
$\Rightarrow \frac{a^2+b^2-c^2}{2ab}=\frac{-1}{2}$
$\Rightarrow \cos C=\frac{-1}{2}$
$\therefore \angle C={120}^0$
and $\sqrt{2}(\sin A+\cos A)=\sqrt{3}$
$\Rightarrow \sqrt{2}\left[\sqrt{2}(\frac{1}{\sqrt{2}}\sin A+\frac{1}{\sqrt{2}}\cos A)\right]=\sqrt{3}$
We know that $\sin \frac{\pi }{4}=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$
$\Rightarrow 2\left[\sin (A+\frac{\pi }{4})\right]=\sqrt{3}$
$\Rightarrow \left[\sin (A+\frac{\pi }{4})\right]=\frac{\sqrt{3}}{2}$
$\Rightarrow \left[\sin (A+\frac{\pi }{4})\right]=\sin \frac{\pi }{3}$
$\Rightarrow A+\frac{\pi }{4}=\frac{\pi }{3}$
$\therefore A=\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}$ is possible.
Option$\left(d\right)$
$\because \sin A+\sin B=\frac{\sqrt{3}+1}{2}$ ................$\left(1\right)$
and $\cos A\cos B=\frac{\sqrt{3}}{4}=\sin A\sin B$
$\therefore \cos A\cos B-\sin A\sin B=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$
$\Rightarrow \cos (A+B)=0$
$\Rightarrow A+B=\frac{\pi }{2}$
$\therefore B=\frac{\pi }{2}-A$
From eqn$\left(1\right)$
$\sin A+\cos A=\frac{\sqrt{3}+1}{2}$
$\Rightarrow \sqrt{2}[\sin A\frac{1}{\sqrt{2}}+\cos A\frac{1}{\sqrt{2}}]=\frac{\sqrt{3}+1}{2}$
We know that $\sin \frac{\pi }{4}=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$
$\Rightarrow \sqrt{2}[\sin A\cos \frac{\pi }{4}+\cos A\sin \frac{\pi }{4}]=\frac{\sqrt{3}+1}{2}$
$\Rightarrow \sin (A+\frac{\pi }{4})=\frac{\sqrt{3}+1}{2\sqrt{2}}$
$\Rightarrow \sin (A+\frac{\pi }{4})=\sin \frac{5\pi }{12}$
$\Rightarrow A+\frac{\pi }{4}=\frac{5\pi }{12}$
$\Rightarrow A=\frac{5\pi }{12}-\frac{\pi }{4}=\frac{\pi }{6}$
$\therefore B=\frac{\pi }{2}-A=\frac{\pi }{2}-\frac{\pi }{6}=\frac{2\pi }{6}=\frac{\pi }{3}$
Thus, $\angle C=\frac{\pi }{2}$ is possible.