How to find the angle between 2 adjacent sloping faces in a right square pyramid?

Question

A right square pyramid vertex O, stands on a square base ABCD. The height is 15 cm and base side length is 10 cm. Find the magnitude of the angle between two adjacent sloping faces. I don't know how to do this. Btw, the answer is 95.74 degrees. I would appreciate any help!

Answer 1

What nonsense is sometimes perpetrated in so-called answers here.



The face of a pyramid is an isosceles triangle. To find the angle between one face and an adjoining face, it is necessary to draw an altitude from one of the lower corners to the opposite sloping edge. Then find the angle which this altitude makes with a similar one on the adjoining face.





As you go through the stuff below, it will be necessary to sketch several triangles as we come to them to make the description clearer (hopefully !)





Take face AOB, and consider the adjoining face DOA.



Call the centre of the base (directly below the vertex) Z, and let M be the mid-point of AB.



Let us establish some other dimensions to begin with.



AB and AD = 10 therefore AM = MB = ZM = 5



OM = √(15² + 5²) = √250 = 5√10



Base diagonal, DB = 10√2



Hence half this diagonal, ZA, ZB, ZD = 5√2



Sloping edges, OA, etc = √(15² + (5√2)²) = √275 = 5√11





In triangleAOB,



Let K be the foot of the perpendicular from B on to the side AO.



And similarly, K will also be the foot of the perpendicular from D on to AO from the adjacent face DOA.



Triangle DKB is an isosceles triangle with base DB, and the vertex angle DKB is the required angle between faces DOA and AOB of the pyramid.



We want to find the length of the sides BK and DK.





In triangle AOM, we have AO = 5√11, AM = 5, OM = 5√10



hence for angle OAM we have sin = OM/OA = √10/√11



. . . . . . . . . . . . . . . . . . . . . . . . . . cos =AM/OA = 1/√11



. . . . . . . . . . . . . . . . . . . . . . . . . . tan = √10





Turning to triangle AKB, we have that



sin angle BAK = BK/AB



and we just found that sin BAK = √10/√11



therefore BK/10 = √10/√11



from which BK = 10√10/√11





Now we have triangle BKD under control.



Base BD = 10√2, sides BK and KD = 10√10/√11



Draw the altitude KZ; and call the angle DKZ, φ



Sin φ = DZ / KD = 5√2 / ( 10√10 /√11) = (5√2 √11) / 10√10 = √11 / √20



φ = sinˉ¹ (√11 / √20) = sinˉ¹ 0.74162 = 47.87º



And the angle we want to find, BKD, = 2φ = 95.74º

Answer 2

This sounds like a 3D problem so remember that the vertex of the pyramid is not over a side but centred over the diagonal of the square base. Since it is a square base the diagonal is the hypotenuse of any 2 adjacent sides, therefore the square root of 10 000. Calculate this length and then take half. The resultant value is the length of the base and you already have the vertical as 15. Using the Pythagorean theorem, you can now calculate the length of the hypotenuse that represents the sloping face of the pyramid, then the angles between the sloping face, the base and the vertical.



In theory, the angle should have been 90 since the interior angles of a square total 360 and a triangle is 180. Review the question, work it out again and then recheck the answer in the back. Be aware, that sometimes the answer in the back is wrong, either a typo or deliberate.

Answer 3

Square Pyramid Angles

Answer 4

Each plane of the pyramid parallel to the base is a square. In my understanding two adjacent sloping faces must have 90° between them. Maybe you should define the expression "adjecent plane" with a drawing.