Bowl of the spherical cap and its diameter Bowl has the shape of spherical cap. Its height is v=15 cm. If you want all the water in the bowl to leak out, you must to tilt the bowl at least 60°.
like this:  

What is the inner diameter of the widest part of the bowl, calculated to the nearest mm?
 A: Nice picture, very helpful for the solution.  Actually, the picture is the solution, it is where the real mathematics is.  Once we see what's happening, the rest is essentially mechanical. We use the tilted version of the picture. 
For $60^\circ$ to be the smallest angle at which all the water leaks out, the line at the bottom must be tangent to the sphere at the point of contact.  Let the point of contact, and tangency, be $T$.  
Draw the vertical line at the point $T$ of tangency. This line goes through the centre $O$ of the sphere. Join $O$ to the bottom point $B$ of the bowl. Suppose that the line $OB$ meets the top surface of the water at $M$. Thus $M$ is the centre of the circle that used to be the top surface of the water. 
Note that $\angle OTM$ is $30$ degrees.  If $R$ is the radius of the complete sphere of which the bowl is a part, then $OM=R-15$.  
The ratio $\frac{R-15}{R}$ is the sine of the $30$ degree angle, so $\frac{R-15}{R}=\frac{1}{2}$, and therefore $R=30$. 
We want the distance $MT$, which is the side of a right-angled triangle with hypotenuse $30$ and other side $15$ (because $30-15=15$). Thus, by the Pythagorean Theorem,  $MT= \sqrt{30^2-15^2}$. Double to get the diameter. Equivalently, we can use the fact that $\frac{MT}{R}$ is the cosine of the $30$ degree angle.
Remark: The angle $60$ is a "nice" angle. That made the calculation easier. However, if it had been the less nice angle of say $61^\circ$, the calculation would have the same structure. Then $\frac{R-15}{R}$ would be the sine of the $29$ degree angle. Solve for $R$, and use the Pythagorean Theorem.   
