The height of the cone SO is 20, the points A and B are on the circumference of the bottom surface, the angle AOB is 90 degrees, and the section SAB forms a dihedral angle of 45 degrees with the bottom surface of the cone. Find the volume of triangular cone S-OAB.

draw

Take the midpoint c of AB and connect SC and OC; Let the radius o of the cone bottom be r.

Because SA=SB, that is, △SAB is an isosceles triangle.

C is the midpoint of AB.

So, SC⊥AB

Also, OA=OB=r, and C is the midpoint of AB.

So, OC⊥AB

So, AB⊥ surface football

So ∠SCO is the angle formed by the surface SAB and the bottom surface, that is ∠ SCO = 45.

Because the bottom of SO⊥ is O, so: △SOC is a right triangle.

So, so =OC=20.

However ∠ AOB = 90, OA=OB.

So △AOB is an isosceles right triangle.

C is the midpoint of the hypotenuse AB.

So △ACO and △BCO are also isosceles right triangles.

Therefore, OA=OB=√2OC=20√2.

Therefore, the area of AOB is s = (1/2) OA * ob = (1/2) * (20 √ 2) 2 = 400.

So the volume of triangular pyramid Vs-A0B = (1/3) SH = (1/3) * 400 * 20 = 8000/3.

The picture is in my space.