Tetrahedral pyramid

Determine the surface of a regular tetrahedral pyramid when its volume is V = 120 and the angle of the sidewall with the base plane is α = 42° 30'.

Correct answer:

S =  200.6434

Step-by-step explanation:

$$\begin{gathered} V=120 \\ A=42+{\displaystyle \frac{30}{60}}={\displaystyle \frac{85}{2}}=42\frac{1}{2}=42.5\text{ ° } \\ {A}_1=A\mathrm{°}\to \text{ rad}=A\mathrm{°}\cdot {\displaystyle \frac{\pi }{180}}=42.5\mathrm{°}\cdot {\displaystyle \frac{3.1415926}{180}}=0.74176=17\pi \mathrm{/}72 \\ \tan A=h\mathrm{/}(a\mathrm{/}2) \\ h=a\mathrm{/}2\cdot \tan A \\ V={\displaystyle \frac{1}{3}}{a}^{2}h \\ V={\displaystyle \frac{1}{6}}{a}^3\tan A \\ a=\sqrt[3]{6\cdot V\mathrm{/}\tan (A_1)}=\sqrt[3]{6\cdot 120\mathrm{/}\tan (0.7418)}\doteq 9.2277 \\ h=a\mathrm{/}2\cdot \tan (A_1)=9.2277\mathrm{/}2\cdot \tan (0.7418)\doteq 4.2278 \\ \sin A_1=h:w \\ w=h\mathrm{/}\sin (A_1)=4.2278\mathrm{/}\sin (0.7418)\doteq 6.258 \\ {S}_1=a\cdot w\mathrm{/}2=9.2277\cdot 6.258\mathrm{/}2\doteq 28.8733 \\ S=a^2+4\cdot S_1=9.2277^2+4\cdot 28.8733=200.6434 \end{gathered}$$

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