Dragan

1)     Quadrilateral  PQRS is an  equilateral quadrilateral  or rhombus.

Tringles  PQR and QSP are identical.

Angle  QSR = 30°    

2)  In the figure below,  AB=BC=CD=DE=EF=FG=GA

What's the question here??? 

Thanks, Lucky!

A rectangular solid, or cuboid, with sides of lengths 2, 10, and 22 is inscribed in a sphere. What is the side length of the cube that can be inscribed in that sphere?

Face diagonal (D_f) of the cuboid's side  10 x  22 is:>     D_f = sqrt(22² + 100²) = 24.16609195

Interior diagonal (D_s) of the cuboid is:>                           D_s = sqrt(D_f² + 2²) = 24.24871131

D_s is the diameter of a sphere and also an interior diagonal of a cube.

The side length of the cube (a) is:>               a = D_s / sqrt(3) = 14    

The area of triangle ABJ =>   A = (3*3)/2 - (3*1)/2 = 3 u²  

∠ABH = ∠HCA = 31°  

   Why is this???

The third angle is 45°        (  180 - 103 - 32 = 45 )

The height of the triangle     h = sqrt(16/2) = 2.828427125

Area of the triangle        A = ( h * 7.3 )/2   =  10.323759 cm²  

QN = 12                PR = 14       PN = 7

F is a foot of the altitude from M to PN

Find angle QPN using QN and PN

Find QP and MP using the Pythagorean theorem

Find MF and PF using angle QPN and MP

Find angle MRF using MF and FR

And finally, find  OR  using NR and angle MRF

Good luck!  

What if it is  150^0 ???  

It's very simple...

Let's find the angle using the tree height and the length of the shadow.

Tree height   H_t = 30 ft

Length of the shadow is 65 ft

Length of the tree shadow is  65 ft - 25 ft = 40 ft

Angle   A = ?                tan(A) = 30/40 ≈ 36.87°

Pole height    H_p = tan(A) * 65 = 48.75 ft