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Question Number 35606 by ajfour last updated on 21/May/18

Commented by ajfour last updated on 21/May/18

$$Hencefindvolumeofthe$$$$triangularpyramid.$$

Commented by MrW3 last updated on 04/Nov/18

$$a=BC=\sqrt{q^2+r^2-2qr\cos \varphi }$$$$b=CA=\sqrt{r^2+p^2-2rp\cos \psi }$$$$c=AB=\sqrt{p^2+q^2-2pq\cos \theta }$$$$\frac{1}{2}{ah}_a=\frac{1}{2}qr\sin \varphi$$$$\Rightarrow h_a=\frac{qr\sin \varphi }{a}=\frac{qr\sin \varphi }{\sqrt{q^2+r^2-2qr\cos \varphi }}$$$$similarly$$$$\Rightarrow h_b=\frac{rp\sin \psi }{\sqrt{r^2+p^2-2rp\cos \psi }}$$$$\Rightarrow h_c=\frac{pq\sin \theta }{\sqrt{p^2+q^2-2pq\cos \theta }}$$$$leth=heightofpyramid$$$$d_a=\sqrt{h_a^2-h^2}$$$$d_b=\sqrt{h_a^2-h^2}$$$$d_c=\sqrt{h_c^2-h^2}$$$$lets=\frac{a+b+c}{2}$$$$A_{base}=\frac{1}{2}({ad}_a+{bd}_b+{cd}_c)=\sqrt{s(s-a)(s-b)(s-c)}$$$$\sqrt{q^2{r}^2{\sin }^2\varphi -a^2{h}^2}+\sqrt{r^2{p}^2{\sin }^2\psi -b^2{h}^3}+\sqrt{p^2{q}^2{\sin }^2\theta -c^2{h}^2}=2\sqrt{s(s-a)(s-b)(s-c)}$$$$\Rightarrow h$$$$V=\frac{h\sqrt{s(s-a)(s-b)(s-c)}}{3}$$

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