Prove that if sin$A < \frac ab$ and $a > b$, then $\angle B$ is acute. Given $\triangle ABC$, $a > b$ and $\angle A$ with the property that sin$A < \frac ab$. How do I prove that $\angle B$ is an acute angle?
I'm  trying to use this and proof that a triangle with this particular property isn't included in the ambiguous case of the sine law.
I'd really appreciate any help.

 A: This is not true. See the diagram below:


Comments on the Edited Question
As mentioned in comments under Pythagoras' answer, it is true that if $a\gt b$, then $A\gt B$. To show this, consider the incircle of our triangle:

Looking at the right triangles $\triangle OEC$ and $\triangle OEB$, we have
$$
\begin{align}
a
&=CE+EB\tag{1a}\\
&=r\cot(C/2)+r\cot(B/2)\tag{1b}
\end{align}
$$
Looking at the right triangles $\triangle ODC$ and $\triangle ODA$, we have
$$
\begin{align}
b
&=CD+DA\tag{2a}\\
&=r\cot(C/2)+r\cot(A/2)\tag{2b}
\end{align}
$$
Thus, if $a\gt b$, then $\cot(B/2)\gt\cot(A/2)$, which, since $\cot(\theta)$ is decreasing on $(0,\pi/2)$, says that $A\gt B$.
Therefore, since the sum of the three angles must be $180^{\large\circ}$, $B$ must be less than $90^{\large\circ}$; otherwise, the sum of $A$ and $B$ would be greater than $180^{\large\circ}$.
A: Your picture is misleading. If $\angle C$ is obtuse, then $\angle B$ is necessarily acute.
In general, your statement is false: Take $a=c=1$ and $b=\sqrt{3}$. Then $\angle A=\angle C=30^\circ$ and $\angle B=120^\circ$ with $$\sin(A)=\frac 1 2<\frac 1{\sqrt{3}}=\frac a b.$$
A: Let $CD\perp AB$ where $D$ lies on $AB$. Then, $\sin A=\frac {CD}{b}<\frac ab\implies CD< a$. Now in $\triangle BCD$, $CD<a\implies \angle B<\angle CDB\implies \angle B <90^{\circ}$ (Because the opposite angle of the larger side is greater than the opposite angle of the smaller side in a triangle). Hence, $\angle B$ is acute.

A: You can use the sine laws to prove this:
From the sine laws we know: $$\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}$$
But we only need: $$\frac{\sin{A}}{a} = \frac{\sin{B}}{b}$$ because side c and angle C are not required for this question.
Then, rearrange the terms to find a similarity with the question: $$\frac{a}{b} = \frac{\sin{A}}{\sin{B}}$$
But, $$\sin{A} < \frac{a}{b}$$ so, $$\sin{A} < \frac{\sin{A}}{\sin{B}}$$
Now remove the like terms,
--> $\sin{A} < \frac{\sin{A}}{\sin{B}}$
--> $1 < \frac{1}{\sin{B}}$
--> $\sin{B} < 1$
