A stationary upright cone has a taper angle `theta=45^@`, and the area of the lateral surface `S_0=4.0m^2`. Find: (a) its taper angle, (b) its lateral surface area, in the reference frame moving with a velocity `v=(4//5)c` along the axis of the cone.

Explanation

In the frame K in which the cone is at rest the coordinates of A are `(0,0,0)` and of B are `(h, h tan theta, 0)`. In the frame `K^'`, which is moving with velocity v along the axis of the cone, the coordinates of A and B at time `t^'` are
`A:(-vt^',0,0), B: (hsqrt(1-beta^2)-vt^', h tan theta, 0)`
Thus the taper angle in the frame `K'` is
`tan theta'=(tan theta)/(sqrt(1-beta^2))(=(y_B'-y_A')/(x_B'-x_A'))`
and the lateral surface area is,
`S=pih'^(2) sec theta' tan theta'`
`=pih^2(1-beta^2)(tantheta)/(sqrt(1-beta^2))sqrt(1+(tan^2theta)/(1-beta^2))=S_0sqrt(1-beta^2cos^2theta)`
Here `S_0=pih^2sectheta tan theta` is the lateral surface area in the rest frame and
`h'=hsqrt(1-beta^2)`, `beta=v//c`.