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. A right angled triangle is possible.

Answer 1

Hint: To solve this question, we will use the two components of the Law of sine or sine rule with sides $b,c$ and angle $\sin B,\sin C$. Using that relation, we will derive the value for the angle $\sin C$. We will then compare the relation with given data and determine the number of triangles which can be made.

Formula Used: The Law of sine:
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

Complete step by step solution: We are given sides $b,c$ and angle $\sin B$ such that $B$ is acute and $bWe will only take the two component of the relationship of Law of sine $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$that is $\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$.
We will now derive the value of angle $\sin C$.
$\sin C=\dfrac{c\sin B}{b}$
As given $b<{c}\sin {B}$ then the value of $\dfrac{c\sin B}{b}$ will be always greater than one. And if $\dfrac{c\sin B}{b}>1$, then $\sin C>1$ which is not possible because we know that the interval of the values for sine is $\left[ -1,1 \right]$. Hence no triangle is possible.

No triangle can be possibly made when $B$ is acute and $b

Note: We must know the interval of the sine because without it we would not be able to solve these kinds of questions. The value $\dfrac{c\sin B}{b}$ will always be greater than one because when we divide a number with a small number then the value is always greater than one.