Question

We are given $b,c$ and $\sin B$ such that $B$ is acute and $b<c\sin B$. Then

• A. No triangle is possible
• B. One triangle is possible
• C. Two triangle are possible
• D. A right angled triangle is possible

Answer 1

Answer: (A)

No triangle is possible

$b,c$ and $\sin B,\angle B$ is acute, $b<c\sin B;$

$\cos B=\frac{a^2+c^2-b^2}{2ac}\Rightarrow a^2-\left(2c\cos B\right)a+(c^2-a^2)=0$

$\Rightarrow D=4c^2{\cos }^{2}B-4(c^2-b^2)=4(b^2-c{\sin }^{2}B)<0$

as $b<c\sin B$

$\Rightarrow a$ is imaginary

$\Rightarrow$ no triangle is possible