Regular quadrangular pyramid

The height of the regular quadrangular pyramid is 6 cm, the length of the base is 4 cm. What is the angle between the ABV and BCV planes?

Result

x =  36.87 °

Solution:

h=6 cm a=4 cm  s1=a/2=4/2=2 cm s2=s12+h2=22+622 10 cm6.3246 cm  tanx/2=s1/h  x=2 180πarctan(s1/h)=2 180πarctan(2/6)36.869936.87365212"h=6 \ \text{cm} \ \\ a=4 \ \text{cm} \ \\ \ \\ s_{1}=a/2=4/2=2 \ \text{cm} \ \\ s_{2}=\sqrt{ s_{1}^2 + h^2 }=\sqrt{ 2^2 + 6^2 } \doteq 2 \ \sqrt{ 10 } \ \text{cm} \doteq 6.3246 \ \text{cm} \ \\ \ \\ \tan x/2=s_{1} / h \ \\ \ \\ x=2 \cdot \ \dfrac{ 180^\circ }{ \pi } \cdot \arctan(s_{1}/h)=2 \cdot \ \dfrac{ 180^\circ }{ \pi } \cdot \arctan(2/6) \doteq 36.8699 \doteq 36.87 ^\circ \doteq 36^\circ 52'12"



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