In a regular 5

In a regular triangular prism ABCV, the angle between the lateral face and the base plane is α = 45°. Determine the angle between the lateral edge and the base plane.

Final Answer:

β =  26.5651 °

Step-by-step explanation:

$$\begin{gathered} \alpha =45{}^{\circ } \\ v=\frac{\sqrt{3}}{2}a \\ \tan \alpha =h:(v/3)=3h/v \\ \tan \alpha =\tan 45°=1 \\ h=v/3=\frac{\sqrt{3}}{6}a \\ s=\sqrt{h^2+(2v/3{)}^2} \\ s=\sqrt{3/36a^2+4/9\cdot 3/4a^2} \\ s=a\sqrt{3/36+1/3} \\ s=a\cdot \sqrt{5/12}=a\cdot \sqrt{\frac{5}{12}} \\ \sin \beta =h:s \\ B=\arcsin (\frac{h}{s}) \\ B=\arcsin (\frac{\frac{\sqrt{3}}{6}a}{\sqrt{\frac{5}{12}}a}) \\ B=\arcsin (\frac{\sqrt{\frac{3}{36}}}{\sqrt{\frac{5}{12}}}) \\ B=\arcsin (\sqrt{\frac{3}{36}\cdot \frac{12}{5}}) \\ B=\arcsin (\sqrt{\frac{3}{36}\cdot \frac{12}{5}}) \\ B=\arcsin (\sqrt{\frac{1}{5}})\doteq 0.4636 \\ \beta =B\to \text{°}=B\cdot \frac{180}{\pi }\text{°}=0.4636\cdot \frac{180}{\pi }\text{°}=26.565\text{°}=26.5651^{\circ }=26°33^{\prime }54" \end{gathered}$$

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