Angle of two lines

There is a regular quadrangular pyramid ABCDV; | AB | = 4 cm; height v = 6 cm. Determine the angles of lines AD and BV.

Result

X =  72.452 °

Solution:

a=4 cm v=6 cm  s=v2+(a/2)2=62+(4/2)2≐2 10 cm≐6.3246 cm  tan⁡X=s/(a/2)  X1=arctan⁡(2⋅ s/a)=arctan⁡(2⋅ 6.3246/4)≐1.2645 rad  X=X1→ ∘=X1⋅ 180π  ∘=72.4516  ∘=72.452∘=72∘27′6"a=4 \ \text{cm} \ \\ v=6 \ \text{cm} \ \\ \ \\ s=\sqrt{ v^2 + (a/2)^2 }=\sqrt{ 6^2 + (4/2)^2 } \doteq 2 \ \sqrt{ 10 } \ \text{cm} \doteq 6.3246 \ \text{cm} \ \\ \ \\ \tan X=s / (a/2) \ \\ \ \\ X_{1}=\arctan( 2 \cdot \ s/a)=\arctan( 2 \cdot \ 6.3246/4) \doteq 1.2645 \ \text{rad} \ \\ \ \\ X=X_{1} \rightarrow \ ^\circ =X_{1} \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =72.4516 \ \ ^\circ =72.452 ^\circ =72^\circ 27'6"

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