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 triangles are possible
• D. one right-angled triangle is posible

Answer 1

Answer: (A)

no triangle is possible

Using cosine rule,

$\cos B=\frac{c^2+a^2-b^2}{2ca}$

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

Now for triangle to exist, discriminant of above quadratic should be non-negative.

$\Rightarrow (2c\cos B{)}^2-4(c^2-b^2)\ge 0\Rightarrow c^2{\cos }^B-c^2+b^2\ge 0$

$\Rightarrow b^2-c^2{\sin }^{2}B\ge 0\Rightarrow c\sin B\le b$

But given $c\sin B>b$.

Hence triangle will not exist.