# Thread: Angle Help Needed thanks!

1. ## Angle Help Needed thanks!

A right square pyramid, vertex O, stands on a square base ABCD. The height is 15cm and base length is 10cm. Find:

a. the length of the slant edge : I got that: 5 root 11.
b. the inclination of a slant edge to the base : 64.76 degrees
c. the inclination of a sloping face to the base : 71.57 degrees

2. Originally Posted by phgao
A right square pyramid, vertex O, stands on a square base ABCD. The height is 15cm and base length is 10cm. Find:

a. the length of the slant edge : I got that: 5 root 11.
b. the inclination of a slant edge to the base : 64.76 degrees
c. the inclination of a sloping face to the base : 71.57 degrees
a.) slant edge,
Slant edge is the hypotenuse of a right triangle whose vertical leg is 15cm and whose horizontal leg is (1/2)[10sqrt(2)] or 5sqrt(2).

So, slant edge = sqrt[15^2 +(5sqrt(2))^2] = sqrt[225 +50] = sqrt[275] = sqrt[25*11] = 5sqrt(11) cm. --------answer.

b.) inclination of slant edge from the base.
Let us call that angle alpha.

sin(alpha) = 15/[5sqrt(11)] = 3/sqrt(11)
alpha = arcsin[3/sqrt(11)] = 64.76 degrees. ------answer.

-------------------------------
c.) inclination of a sloping face from the base.
Let us call that angle beta.

Imagine a right triangle, with these:
---hypotenuse = sloping face
---vertical leg = 15cm
---horizontal leg = 10/2 = 5cm
---angle between hypotenuse and horizontal leg = angle beta.

tan(beta) = 15/5 = 3
beta = arctan(3) = 71.565 degrees. -------answer.

-----------------------------------------
d.) magnitude of angle between two adjacent sloping faces.
Let us call that angle gamma.

In the right triangle where a slant edge is the hypotenuse, project a line segment from the intersection of the vertical and horizontal legs (actually, the center of the square base of the pyramid) to the slant edge, such that this line segment is perpendicular to the slant edge. Let us call this line segment, x. Line segment x intersects the slant edge at, say, point E.
Another right triangle is formed, with these:
--hypotenuse = 5sqrt(2) cm -------half of diagonal.
--one leg = x
--the other leg = unknown ----a portion of the slant edge, say, line segment DE, where DO is the slant edge.
--angle opposite leg x is 64.76 degrees.

sin(64.76deg) = x/5sqr(2)
x = [5sqrt(2)]sin(64.76deg) = 6.396 cm.

Now, in the whole pyramid, project a line segment from a corner of the square base, say, corner A, to point E.
Yet another right triangle is formed, with these:
---hypotenuse = AE
---one leg = x = 6.396 cm
---the other leg = 5sqrt(2) ---------half of diagonal AC of the square base.
---angle between x and hypotenuse is half of gamma.

tan(gamma/2) = [5sqrt(2)]/6.396 = 1.105545
gamma/2 = arctan(1.105545) = 47.87 degrees.
Therefore, gamma = 2(47.87deg) = 95.74 degrees --------answer.

3. Hey thanks so much ticbol! You are the best and provide detailed clear answers + explanations! You rock!

4. How come when you work out the answer (d) this way it is different? Help!

: First drop the perpendicular from O - the top to the front line AB. From this we work out angleOBA is 72.45 degrees.

: Say A is the front right corner of pyrimid, then drop a perpendicular to line OB where O is the upper vertex and B is the front left had corner.

: Look at this right angled triangle; we already know OBA - 72.45, and we know AB - 10cm. So we can work out the line say Q from A perpendicular to the point on OB say P.

:So Q = 9.5346cm

:Then join points A and C (back right corner), join C to P, and join P to A. We have a triangle. Let angle APC = F which is the angle we want.

:So cosine rule:

c^2 = a^2 + b^2 - 2ab(cosF)
cosF = a^2+b^2-c^2/2ab
cosF = 2*9.5346^2-200/2*9.5346^2
...
F = 118.47 degrees?! Where is the mistake if there is one?

5. Originally Posted by phgao
How come when you work out the answer (d) this way it is different? Help!

: First drop the perpendicular from O - the top to the front line AB. From this we work out angleOBA is 72.45 degrees.

: Say A is the front right corner of pyrimid, then drop a perpendicular to line OB where O is the upper vertex and B is the front left had corner.

: Look at this right angled triangle; we already know OBA - 72.45, and we know AB - 10cm. So we can work out the line say Q from A perpendicular to the point on OB say P.

:So Q = 9.5346cm

:Then join points A and C (back right corner), join C to P, and join P to A. We have a triangle. Let angle APC = F which is the angle we want.

:So cosine rule:

c^2 = a^2 + b^2 - 2ab(cosF)
cosF = a^2+b^2-c^2/2ab
cosF = 2*9.5346^2-200/2*9.5346^2
...
F = 118.47 degrees?! Where is the mistake if there is one?
Very good! You're way is correct, except that you made a mistake in your computations (arithmetic or algebra or use of calculator) using the Law of Cosines.

c^2 = a^2 +b^2 -2ab*cosF
cosF = (a^2 +b^2 -c^2)/(2ab)
cosF = [(9.5346)^2 +(9.5346)^2 -(10sqrt(2))^2] / [2(9.5346)(9.5346)]
cosF = [2(9.5346)^2 -200] / [2(9.5346)^2]
cosF = (-18.1828)/(181.8172)
cosF = -0.10000594
F = arccos(-0.10000594) = 95.7395 degrees, not 118.47deg.

6. Oh ok, thanks a lot.