Answer:
The angle formed between CF and the plane ABCD is approximately 47.14°
Step-by-step explanation:
The given parameters are;
BC = 6.8
DE = 9.3
∠BAC = 52°
We note that the angles formed by the vertex of a cuboid are right triangles, therefore, by trigonometric ratios, we get;
sin∠BAC = BC/(The length of a line drawn from A to C)
∴ The length of the line drawn from A to C = BC/sin∠BAC
The length of the line drawn from A to C = 6.8/sin(52°) ≈ 8.63
∴ AC = 8.63
By trigonometry, we have;
The angle formed between CF and the plane ABCD = Angle ∠ACF
[tex]tan\angle X = \dfrac{Opposite \ leg \ length}{Adjacent\ leg \ length}[/tex]
[tex]tan\angle ACF = \dfrac{FA}{AC}[/tex]
In a cuboid, FA = BG = CH = DE = 9.3
[tex]\therefore tan\angle ACF = \dfrac{9.3}{8.63}[/tex]
[tex]\therefore \angle ACF = arctan \left(\dfrac{9.3}{8.63} \right) \approx 47.14^{\circ}[/tex]
The angle formed between CF and the plane ABCD = Angle ∠ACF ≈ 47.14°
