![\bf \textit{Law of sines} \\\\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{sin(100^o)}{2.5}=\cfrac{sin(40^o)}{r}\implies r\cdot sin(100^o)=2.5\cdot sin(40^o) \\\\\\ r=\cfrac{2.5\cdot sin(40^o)}{sin(100^o)}\implies \boxed{r\approx 1.63}](/tpl/images/0404/7704/b7cff.png)
![\bf \cfrac{r}{sin(35^o)}=\cfrac{h}{sin(55^o)}\implies r\cdot sin(55^o)=h\cdot sin(35^o) \\\\\\ \cfrac{r\cdot sin(55^o)}{sin(35^o)}=h\implies \cfrac{1.63\cdot sin(55^o)}{sin(35^o)}\approx h\implies \boxed{2.33\approx h} \\\\[-0.35em] ~\dotfill\\\\ \textit{volume of a circular cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r\approx 1.63\\ h\approx 2.33 \end{cases}\implies V\approx \cfrac{\pi (1.63)^2(2.33)}{3}\implies \blacktriangleright V\approx 6.48 \blacktriangleleft](/tpl/images/0404/7704/24a79.png)
Can someone find the volume of this pyramid?
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