Quadrilateral oblique prism

What is the volume of a quadrilateral oblique prism with base edges of length a = 1m, b = 1.1m, c = 1.2m, d = 0.7m, if a side edge of length h = 3.9m has a deviation from the base of 20° 35 ´ and the edges a, b form an angle of 50.5°.

Correct result:

V =  1.0113 m³

Solution:

$$\begin{gathered} a=1\text{ m } \\ b=1.1\text{ m } \\ c=1.2\text{ m } \\ d=0.7\text{ m } \\ h=3.9\text{ m } \\ \phi =20+35\mathrm{/}60={\displaystyle \frac{247}{12}}\doteq 20.5833^{\circ } \\ \beta =50.5^{\circ } \\ {u}^2=a^2+b^2-2\cdot a\cdot b\cdot \cos \beta \\ u=\sqrt{a^2+b^2-2\cdot a\cdot b\cdot \cos {\beta }^{\circ }}=\sqrt{a^2+b^2-2\cdot a\cdot b\cdot \cos 50.5^{\circ }}=\sqrt{1^2+1.1^2-2\cdot 1\cdot 1.1\cdot \cos 50.5^{\circ }}=\sqrt{1^2+1.1^2-2\cdot 1\cdot 1.1\cdot 0.636078}=0.90035 \\ s=(a+b+u)\mathrm{/}2=(1+1.1+0.9003)\mathrm{/}2\doteq 1.5002\text{ m } \\ {S}_1=\sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-u)}=\sqrt{1.5002\cdot (1.5002-1)\cdot (1.5002-1.1)\cdot (1.5002-0.9003)}\doteq 0.4244\text{ m } \\ {s}_2=(c+d+u)\mathrm{/}2=(1.2+0.7+0.9003)\mathrm{/}2\doteq 1.4002\text{ m } \\ {S}_2=\sqrt{s_2\cdot (s_2-c)\cdot (s_2-d)\cdot (s_2-u)}=\sqrt{1.4002\cdot (1.4002-1.2)\cdot (1.4002-0.7)\cdot (1.4002-0.9003)}\doteq 0.3132\text{ m } \\ {h}_2=h\cdot \sin {\phi }^{\circ }=h\cdot \sin 20.5833333333^{\circ }=3.9\cdot \sin 20.5833333333^{\circ }=3.9\cdot 0.351569=1.37112 \\ S=S_1+S_2=0.4244+0.3132\doteq 0.7376\text{ m } \\ V=S\cdot h_2=0.7376\cdot 1.3711=1.0113{\text{m}}^3 \end{gathered}$$

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