Square pyramid

Calculate the volume of the pyramid with the side 5cm long and with a square base, side-base has angle of 60 degrees.

Result

V =  18.042 cm3

Solution:

s=5 cm A=60 ∘ cos⁡(A)=(u/2)/s cos⁡(A)=u/(2s) u=2⋅ s⋅ cos⁡(A∘→ rad)=2⋅ s⋅ cos⁡(A∘⋅ π180 )=2⋅ 5⋅ cos⁡(60∘⋅ 3.1415926180 )=5 a=u/2=5/2≐3.5355 cm S=a2=3.53552=252=12.5 cm2 h=s⋅ sin⁡(A∘→ rad)=s⋅ sin⁡(A∘⋅ π180 )=5⋅ sin⁡(60∘⋅ 3.1415926180 )=4.33013 V=S⋅ h/3=12.5⋅ 4.3301/3≐18.0422≐18.042 cm3s=5 \ \text{cm} \ \\ A=60 \ ^\circ \ \\ \cos(A)=(u/2) / s \ \\ \cos(A)=u/(2s) \ \\ u=2 \cdot \ s \cdot \ \cos( A ^\circ \rightarrow\ \text{rad} )=2 \cdot \ s \cdot \ \cos( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=2 \cdot \ 5 \cdot \ \cos( 60 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=5 \ \\ a=u / \sqrt{ 2 }=5 / \sqrt{ 2 } \doteq 3.5355 \ \text{cm} \ \\ S=a^2=3.5355^2=\dfrac{ 25 }{ 2 }=12.5 \ \text{cm}^2 \ \\ h=s \cdot \ \sin( A ^\circ \rightarrow\ \text{rad} )=s \cdot \ \sin( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=5 \cdot \ \sin( 60 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=4.33013 \ \\ V=S \cdot \ h / 3=12.5 \cdot \ 4.3301 / 3 \doteq 18.0422 \doteq 18.042 \ \text{cm}^3

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