6 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
a the length of the slant edge b the inclination of a slant edge to the base
c the inclination of a sloping face to the base
d the magnitude of the angle between two adjacent sloping faces.
6A
assuming you meant the slant edge OA (or OB or OC or OD)
you can construct a right angled triangle that cuts the pyramid from the centre to the edge (across the diagonals)
let X be the centre of the base
^2+15^2}=5\sqrt{11})
=sin^{-1}\left(\frac{15}{5\sqrt{11}}\right)=1.13^c=64.8^o)
6C
to do this, we have to find the magnitude of centre of the slant face
we can find this by constructing a right-angled triangle on the slant face, with the hypotenuse being OA (or OB or OC or OD)
let M be the midpoint of AB
 ^2 - 5^2 }=5\sqrt{10})
the angle the slant face makes with the base is then
=sin^{-1}\left( \frac{15}{5\sqrt{10}} \right) = 1.25^c=71.6^o)
6D
the approach to this problem is not obvious at the start
but we do know that:
1.the angle is measured on a plane perpendicular to the slant edge (can someone prove that 
)2.since the slant edge is slanting "in", the plane will cut "downwards" (not the most pedantic definition, but its easier to visualize)
hence, if we imagine perpendicularly cutting the slant OB at a point P such that it ends on the diagonal AC, we'll have an isoceles triangle (
ACP where AC goes through X)
then the angle we need will simply be
APC, or 2
APX
to do this, however, we must first find the magnitude of AP (perpendicular to OB)
-this requires us to first find
 = cos^{-1}\left( \frac{5}{5\sqrt{11}} \right)=1.26^c=72.5^o)
Then in our
APB
=10sin(72.5^o)=10\sqrt{\frac{10}{11}})
with this we can now construct our final
APC (
APX) given |AP| and |AX|, and find out angle
 = 2\cdot sin^{-1} \left( \frac{5\sqrt{2}}{10\sqrt{\frac{10}{11}}} \right) = 2\cdot sin^{-1}\left( \frac{\sqrt{55}}{10} \right) = 1.67^c = 95.7^o)
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