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.

Correct answer:

X =  72.4516 °

Step-by-step explanation:

$$\begin{gathered} a=4\text{ cm } \\ v=6\text{ cm } \\ s=\sqrt{v^2+(a\mathrm{/}2{)}^2}=\sqrt{6^2+(4\mathrm{/}2{)}^2}=2\sqrt{10}\text{ cm}\doteq 6.3246\text{ cm } \\ \tan X=s\mathrm{/}(a\mathrm{/}2) \\ {X}_1=\arctan (2\cdot s\mathrm{/}a)=\arctan (2\cdot 6.3246\mathrm{/}4)\doteq 1.2645\text{ rad } \\ X=X_1{\to }^{\circ }=X_1\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=1.2645\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=72.452^{\circ }=72.4516^{\circ }=72^{\circ }27^{\mathrm{\prime }}6\mathrm{"} \end{gathered}$$

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