Hello, Ilsa!

Consider a right circular cone with the vertex $\displaystyle V$ and the center of the base $\displaystyle O$.

The radius of the base is 1 and the height $\displaystyle VO$ is $\displaystyle \sqrt{35}$

Consider also the diameter $\displaystyle AB$ of the base and $\displaystyle C$ the midpoint of $\displaystyle VA$.

Compute the length of the shortest path, on the surface of the cone from $\displaystyle B$ to $\displaystyle C.$ Code:

V
*
/|\
/ | \
/ | \
C * __| \ 6
/ √35| \
/ | \
/ | \
A * - - - * - - - * B
O 1

We have a right triangle with legs 1 and $\displaystyle \sqrt{35}.$

The hypotenuse (slant height of the cone) is 6.

If we "unroll" *half* the cone and press it flat,

. . we have the diagram below.

Code:

V
*
/ \
/30°\ 3
/ \ C
6 / *
/ * \
/ * \ 3
/ * \
B * - - - - - - - *
* * *
π

The circumference of the base of the cone is: .$\displaystyle 2\pi r \:=\:2\pi(1) \:=\:2\pi$

Half the circumference is: $\displaystyle \pi$

We have: .$\displaystyle s \:=\:r\theta \quad\Rightarrow\quad \theta \:=\:\frac{s}{r}$

. . Hence: .$\displaystyle \theta \:=\:\frac{\pi}{6} $

The vertex angle is 30°.

The shortest distance between $\displaystyle B$ and $\displaystyle C$ is a straight line.

In $\displaystyle \Delta BVC$, Law of Cosines:

. . $\displaystyle BC^2 \:=\:6^2 + 3^2 - 2(6)(3)\cos30^o \;=\;45 - 36\left(\frac{\sqrt{3}}{2}\right) \;=\;9(5 - 2\sqrt{3}) $

Therefore: .$\displaystyle BC \;=\;3\sqrt{5-2\sqrt{3}} \;\approx\;3.718 $