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
![tan\angle X = \dfrac{Opposite \ leg \ length}{Adjacent\ leg \ length}](https://tex.z-dn.net/?f=tan%5Cangle%20X%20%3D%20%5Cdfrac%7BOpposite%20%5C%20leg%20%5C%20length%7D%7BAdjacent%5C%20leg%20%5C%20length%7D)
![tan\angle ACF = \dfrac{FA}{AC}](https://tex.z-dn.net/?f=tan%5Cangle%20ACF%20%3D%20%5Cdfrac%7BFA%7D%7BAC%7D)
In a cuboid, FA = BG = CH = DE = 9.3
![\therefore tan\angle ACF = \dfrac{9.3}{8.63}](https://tex.z-dn.net/?f=%5Ctherefore%20tan%5Cangle%20ACF%20%3D%20%5Cdfrac%7B9.3%7D%7B8.63%7D)
![\therefore \angle ACF = arctan \left(\dfrac{9.3}{8.63} \right) \approx 47.14^{\circ}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cangle%20ACF%20%3D%20arctan%20%5Cleft%28%5Cdfrac%7B9.3%7D%7B8.63%7D%20%5Cright%29%20%5Capprox%2047.14%5E%7B%5Ccirc%7D)
The angle formed between CF and the plane ABCD = Angle ∠ACF ≈ 47.14°
Work out 6% of $42 which equals to $2.52. The work out what 20% of $42 which is $8.40. Finally add the $8.40 and $2.52 onto $42 to give you your answer of $52.92
Rational radical 2 times radical 8 is radical 16 which is 4
Answer: That is a right triangle sir
Step-by-step explanation:
You see the box in the corner of the triangle. Well it's not just a box that box measures 90 degrees and a right triangle is 90 degrees so there is your answer.