Quadrilateral pyramid

The regular quadrilateral pyramid has a base edge a = 1.56 dm and a height h = 2.05 dm.

Calculate:

a) the deviation angle of the sidewall plane from the base plane
b) deviation angle of the side edge from the plane of the base

Correct answer:

A =  69.1688 °
B =  61.7157 °

Step-by-step explanation:

$$\begin{gathered} a=1.56\text{dm} \\ h=2.05\text{dm} \\ \tan A=h:(a/2) \\ A=\frac{180°}{\pi }\cdot \arctan (2\cdot h/a)=\frac{180°}{\pi }\cdot \arctan (2\cdot 2.05/1.56)=69.1688^{\circ }=69°10^{\prime }8" \end{gathered}$$

$$\begin{gathered} u=\sqrt{2}\cdot a=\sqrt{2}\cdot 1.56\doteq 2.2062\text{dm} \\ \tan B=h:(u/2) \\ B=\frac{180°}{\pi }\cdot \arctan (2\cdot h/u)=\frac{180°}{\pi }\cdot \arctan (2\cdot 2.05/2.2062)=61.7157^{\circ }=61°42^{\prime }57" \end{gathered}$$

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