1. ## solved

solved

2. Hello, xMichaelx!

A Cone has a slant height $m$ and a semi-vertical angle $2a$

Prove that the radius $r$ of the inscribed sphere is given by:

. . $r\:=\: \frac {m\sin 4a}{2(1+\sin 2a)}$
Code:
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/|\
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/  |2a\
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/   * * *   \
/*     |     *\  E
*       |       o
m    *        |     *  *    m
/         |   * r   \
/*         | *       *\
/ *       P o         * \
/  *         |         *  \
/             |             \
/     *        |r       *     \
/       *       |       *       \
/          *     |     *          \
B o - - - - - - - * o * - - - - - - - o C
D

The side view of the cone is $\Delta ABC\!:\;AB = AC = m$
P is the center of the inscribed sphere: . $PD = PE = r$
Note that $\angle PEA = 90^o$

In right triangle $AEP\!:\;\sin2a \:=\:\frac{r}{AP} \quad\Rightarrow\quad AP \:=\:\frac{r}{\sin2a}$

. . Hence: . $AD \:=\:AP + PD \:=\:\frac{r}{\sin2a} + r \quad\Rightarrow\quad AD \:=\:r\left(\frac{1+\sin2a}{\sin2a}\right)$ .[1]

In right triangle $ADC\!:\:\cos2a \:=\:\frac{AD}{m} \quad\Rightarrow\quad AD \:=\:m\cos2a$ .[2]

Equate [1] and [2]: . $r\left(\frac{1+\sin2a}{\sin2a}\right) \:=\:m\cos2a \quad\Rightarrow\quad r \:=\:\frac{m\sin2a\cos2a}{1+\sin2a}$

Therefore: . $r \:=\:\frac{m({\color{red}2}\sin2a\cos2a)}{{\color{ red}2}(1+\sin2a)} \quad\Rightarrow\quad r \;=\;\frac{m\sin4a}{2(1+\sin2a)}$