Deviation of the lines

Find the deviation of the lines AG, BH in the ABCDEFGH box-cuboid, if given | AB | = 3cm, | AD | = 2cm, | AE | = 4cm

Correct result:

A =  33.855 °

Solution:

a=3 cm b=2 cm c=4 cm  u1=b2+c2=22+422 5 cm4.4721 cm u2=a2+u12=32+4.4721229 cm5.3852 cm  sinA=au2  A=180πarcsin(au2)=180πarcsin(35.3852)=33.855=335116"a=3 \ \text{cm} \ \\ b=2 \ \text{cm} \ \\ c=4 \ \text{cm} \ \\ \ \\ u_{1}=\sqrt{ b^2+c^2 }=\sqrt{ 2^2+4^2 } \doteq 2 \ \sqrt{ 5 } \ \text{cm} \doteq 4.4721 \ \text{cm} \ \\ u_{2}=\sqrt{ a^2 + u_{1}^2 }=\sqrt{ 3^2 + 4.4721^2 } \doteq \sqrt{ 29 } \ \text{cm} \doteq 5.3852 \ \text{cm} \ \\ \ \\ \sin A=\dfrac{ a }{ u_{2} } \ \\ \ \\ A=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin (\dfrac{ a }{ u_{2} } )=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin (\dfrac{ 3 }{ 5.3852 } )=33.855 ^\circ =33^\circ 51'16"



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