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3 years ago

Cuboid ABCDEFGH is shown

Mathematics

1 answer:

gtnhenbr [62]3 years ago

4 0

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

$tan\angle X = \dfrac{Opposite \ leg \ length}{Adjacent\ leg \ length}$

$tan\angle ACF = \dfrac{FA}{AC}$

In a cuboid, FA = BG = CH = DE = 9.3

$\therefore tan\angle ACF = \dfrac{9.3}{8.63}$

$\therefore \angle ACF = arctan \left(\dfrac{9.3}{8.63} \right) \approx 47.14^{\circ}$

The angle formed between CF and the plane ABCD = Angle ∠ACF ≈ 47.14°

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