The lateral edge of a regular quadrangular pyramid is 10 cm and forms an angle of 30
| September 25, 2021 | education
The lateral edge of a regular quadrangular pyramid is 10 cm and forms an angle of 30 ° with its height. Find the linear angle of the dihedral at the base.
In a right-angled triangle MOB, the leg OB lies opposite the angle 30, then OB = BM / 2 = 10/2 = 5 cm.
ОМ ^ 2 = ВМ ^ 2 – ОВ ^ 2 = 100 – 25 = 75.
ОМ = 5 * √3 cm.
Since ABCD is a square, then BD = 2 * OB = 2 * 5 = 10 cm.
Then BD ^ 2 = AB ^ 2 + AD ^ 2 = 2 * AB ^ 2.
100 = 2 * AB ^ 2.
AB ^ 2 = 100/2 = 50.
AB = 5 * √2 cm.
The OH segment is equal to half the length of the side of the square. OH = AD / 2 = 5 * √2 / 2.
In a right-angled triangle MOH, tgOHM = OM / OH = 5 * √3 / (5 * √2 / 2) = 2 * √3 / √2 = √6.
Angle OHM = arctan (√6).
Answer: The linear angle of the dihedral at the base is arctan (√6).
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