Side deviation

Frustum has the base radii of the figures r1 and r2: r1> r2, r2 = s, and if the side deviation from the base plane is 60°. Express the surface and volume of the cone frustum using its side s.

Correct answer:

S = 18.0642 s^2
V = 4.3078 s^3

Step-by-step explanation:

$$\begin{gathered} r_1>r_2 \\ {r}_2=s \\ \alpha =60°\to \text{rad}=60°\cdot \frac{\pi }{180}=60°\cdot \frac{3.1415926}{180}=1.0472=\pi /3 \\ {r}_1=s+s\cdot \cos \alpha \\ {k}_2=1+\cos \alpha =1+\cos 1.0472=\frac{3}{2}=1\frac{1}{2}=1.5 \\ {r}_1=1.5s \\ \sin \alpha =h:s \\ h=s\cdot \sin \alpha \\ {k}_1=\sin \alpha =\sin 1.0472\doteq 0.866 \\ h=k_1\cdot s \\ S=\pi (r_1^2+r_2^2)+\pi (r_1+r_2).s \\ {k}_3=\pi \cdot (1.5^2+1^2)+\pi \cdot (1.5+1)=3.1416\cdot (1.5^2+1^2)+3.1416\cdot (1.5+1)\doteq 18.0642 \\ S=k_3\cdot s^2 \\ S=18.0642s^2 \end{gathered}$$

$$\begin{gathered} V=\frac{1}{3}.\pi .h.(r_1^2+r_1.r_2+r_2^2) \\ {k}_4=\frac{1}{3}\cdot \pi \cdot k_1\cdot (1.5^2+1.5\cdot 1+1^2)=\frac{1}{3}\cdot 3.1416\cdot 0.866\cdot (1.5^2+1.5\cdot 1+1^2)\doteq 4.3078 \\ V=k_4{s}^3 \\ V=4.3078s^3 \end{gathered}$$

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