Question

Can there exist triangles ABC satisfying the following relations? Write yes or no giving reasons:
(i) $tanA+tanB+tanC=0$
(ii) $\frac{sinA}{2}=\frac{sinB}{3}=\frac{sinC}{7}$
(iii) $(a+b{)}^2=c^2+ab$ and
$\sqrt{2}(sinA+cosA)=\sqrt{3}$
(iv) $sinA+sinB=\frac{\sqrt{3}+1}{2}$
$cosAcosB=\frac{\sqrt{3}}{4}=sinAsinB$

Answer 1

(i) No. Since $A+B+C=\pi$
$\therefore \sum tanA=tanAtanBtanC=0$
$\Rightarrow$ Either A = 0 or B = 0 or C = 0
which is not possible in a triangle.
(ii) No. By sine rule, $\frac{a}{2}=\frac{b}{3}=\frac{c}{7}=\frac{a+b}{5}$
$\therefore a+b=\frac{5}{7}c<c$. Not possible as sum of two sides is always greater than the third.
(iii) Yes.
$a^2+b^2-c^2+ab=0\Rightarrow 2abcosC+ab=0$
$cosC=-\frac{1}{2}\Rightarrow C={120}^o$
Also, from 2nd relation
We have $sin(A+\frac{\pi }{4})=\frac{\sqrt{3}}{2}=sin\frac{\pi }{3}=\therefore A={15}^o$
and hence $B={45}^o$. Such a $\Delta$ is possible.
(iv) Yes. Adding and subtracting the relation in 2nd,
$cos(A-B)=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\therefore A-B={30}^o$
$cos(A+B)=0$ $\therefore A+B={90}^o$
$\therefore A={60}^o,B={30}^o$
$\therefore C={180}^o-(A+B)={90}^o$
These values also satisfy the first given relation.