Answer:
As follows,
Step-by-step explanation:
Find the attachment.
Distance of PQ=√13
Distance of QR=√13
Distance of PR=√26
Thus, PQR is an isosceles triangle.
Distance of MN=√(16+(a-2)^2)
16=16+a^2-4a+4
a^2-4a+4=0
a=2
Distance of RS=√(9+(b+1)^2)
25=9+b^2+2b+1
b=3 or -5
Distance of RS=√(9+(-1-c)^2)
18=9+c^2 + 2 c + 1
c=-4,2
Answer:
The angle between [A_F] and the base of the cone = 68.2°
The area of the base of the cone ≈ 12.57 m²
Step-by-step explanation:
The given parameters are;
The height of the cone = 5 m
The base radius of the cone = 2 m
The angle which the AC = 120°
Therefore, we have;
The angle between [A_F] and the base of the cone = The angle between [CF] and the base of the cone
The angle between [CF] and the base of the cone = tan⁻¹(5/2) = tan⁻¹(2.5) ≈ 68.2°
∴ The angle between [A_F] and the base of the cone = The angle between [CF] and the base of the cone = 68.2°
The angle between [A_F] and the base of the cone = 68.2°
The area of the base of the cone = π × r² = π × 2² = 4·π ≈ 12.57
The area of the base of the cone ≈ 12.57 m².