Through the center O of the square ABCD, the perpendicular OF is drawn to the plane of the square.

| June 21, 2021 | education

Through the center O of the square ABCD, the perpendicular OF is drawn to the plane of the square. Find the angle between planes BCF and ABCD if FB = 5, BC = 6.

Let us draw a perpendicular from point O to the side of the square AB. Since ABSD is a square, then the triangle of the COB is isosceles, then the height OH is also the median of the angle BOC and therefore BH = CH = HO = BC / 2 = 6/2 = 3 cm.

Consider a right-angled triangle BHF, in which the angle H is straight, BH = 3 cm, BF = 5 cm, then by the Pythagorean theorem:

FH ^ 2 = BF ^ 2 – BH ^ 2 = 25 – 9 = 16 cm.

FH = 4 cm.

Consider a right-angled triangle FHO, which has a straight line angle O, hypotenuse FH = 4 cm, leg OH = 3 cm, then:

Cos∠FHO = OH / FH = 3/4.

∠FHO = arccos3 / 4 ≈ 41.

Answer: ∠FHO = ≈ 41.



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