Space diagonal angles

Calculate the angle between the body diagonal and the side edge c of the block with dimensions: a = 28cm, b = 45cm and c = 73cm. Then, find the angle between the body diagonal and the plane of the base ABCD.

Correct result:

C =  35.9807 °
A =  54.0193 °

Solution:

$$\begin{gathered} a=28\text{ cm } \\ b=45\text{ cm } \\ c=73\text{ cm } \\ {u}_1=\sqrt{a^2+b^2}=\sqrt{28^2+45^2}=53\text{ cm } \\ \tan C=u_1:c \\ C={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan (u_1\mathrm{/}c)={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan (53\mathrm{/}73)=35.9807^{\circ }=35^{\circ }58^{\mathrm{\prime }}50\mathrm{"} \end{gathered}$$

$$\begin{gathered} A+C=90^{\circ } \\ A=90-C=90-35.9807=54.0193^{\circ }=54^{\circ }{1}^{\mathrm{\prime }}10\mathrm{"} \end{gathered}$$

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