The figure above shows a solid. The cross-section of the solid is a sector of a circle of radius 5 cm and angle 30°. The horizontal cross-sections, OST and RUV, are 7 cm apart. S, T, U and V lie on the curved surface of the solid. The lines OR, TU and SV are vertical.

1. Find the arc length of the sector forming the curved edge of the surface.

Arc length
$=\frac{30}{360}\times 2\pi(5)$
$=\frac{5\pi}{6}\text{ cm}$

2. Multiply the arc length by the vertical distance to get the curved surface area.

Curved surface area
$=\frac{5\pi}{6}\times 7$
$=\frac{35\pi}{6}\text{ cm}^2$

1. Find the straight-line length $ST$ of the chord in the sector.

In the horizontal cross-section $OST$, $OS=OT=5$ cm and $\angle SOT=30^\circ$.

Using the cosine rule in $\triangle OST$,
$ST^2=5^2+5^2-2(5)(5)\cos30^\circ$
$ST=\sqrt{50-50\cos30^\circ}=2.588\ldots\text{ cm}$

2. Use right-angled trigonometry in triangle $STU$ to find $\angle UST$.

Since $TU$ is vertical and $ST$ lies in the horizontal plane, $TU\perp ST$.

In right-angled $\triangle STU$, $TU=7$ cm.
$\tan \angle UST=\frac{TU}{ST}=\frac{7}{2.588\ldots}$
$\angle UST=69.7^\circ\text{ }(3\text{ s.f.})$

1. Cube the ratio of corresponding radii to obtain the volume ratio.

Smaller radius : larger radius
$=3:5$
For similar solids,
$\text{volume ratio}=(\text{length ratio})^3$
So
$\text{smaller volume : larger volume}=3^3:5^3=27:125.$