# 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:

$a = 4 \ cm \ \\ v = 6 \ cm \ \\ \ \\ s = \sqrt{ v^2 + (a/2)^2 } = \sqrt{ 6^2 + (4/2)^2 } = 2 \ \sqrt{ 10 } \ cm \doteq 6.3246 \ cm \ \\ \ \\ \tan X = s / (a/2) \ \\ \ \\ X_{ 1 } = \arctan( 2 \cdot \ s/a) = \arctan( 2 \cdot \ 6.3246/4) \doteq 1.2645 \ rad \ \\ \ \\ X = X_{ 1 } \rightarrow \ ^\circ = X_{ 1 } \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ = 72.4515993865 \ \ ^\circ = 72.452 ^\circ = 72^\circ 27'6"$

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