Get Going

$\sin ABP$ $= \frac {AP * sin 40}{BP}$ $= \frac {2.532 CB * sin 40}{\sqrt{3} CB}$

$\because$ $\sin \theta = \sin (180 - \theta)$ $\Rightarrow$ $\angle ABP=70^\circ\ or =110^\circ$

Then we must test which one satisfy other relations in the problem.

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Proof:

The triangle inequality states that the sum of the lengths of any two sides of a triangle is greater than the length of the remaining side.

In $\triangle ABC$

$\Rightarrow$ $AB<AC+CB$ $\dashrightarrow 1$

$\angle ABP=70^\circ\ or =110^\circ$ If $\angle ABP=70^\circ$ $\Rightarrow$ $\angle P=70^\circ$

$\Rightarrow$ $\triangle ABP$ is isosceles triangle $AB=AP$ .

$\therefore$ $AB=AP=AC+2 CB$ $\dashrightarrow 2$

But $2$ contradicts $1$ $\Rightarrow$ $\angle ABP=110^\circ$

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Another Proof:

In $\triangle CBP$

$CP>CB$ $\Rightarrow$ $\angle CBP > \angle P$ $\dashrightarrow 1$

If $\angle ABP=70^\circ$ $\Rightarrow$ $\angle CBP=50^\circ$ , $\angle P=70^\circ$ $\Rightarrow$ $\angle CBP < \angle P$ $\dashrightarrow 2$

But $2$ contradicts $1$ $\Rightarrow$ $\angle ABP=110^\circ$

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Hint

After you get that $BP = \sqrt{3} CB$

As $CB = 1 CB$ , $BP = \sqrt{3} CB$ , $CP = 2 CB$ $\Rightarrow$ Ratio $1:\sqrt{3}:2$

$\Rightarrow$ $\triangle BCP$ is a $30-60-90$ triangle $\Rightarrow$ $\angle CBP=90^\circ$

$\Rightarrow$ $\angle ABP= 20 + 90 =110^\circ$

If you draw the diagram then you would realize that you obtained the reference angle. The real angle is 180 - 70 which equals 110.