# Maths Help - 3D Trig (1 Viewer)

#### 1729

##### Active Member

I need help with question b), I've tried many variations but I couldn't get the correct answer but I'm probably overthinking it, (i just need to know the angle).

Thanks for the help.
$\bg_white \noindent If AC is the diagonal of the square base ABCD and O is the vertex of the pyramid and AP is drawn such that AP \perp OB then the required angle is \angle APC.$

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#### Andy005

##### Member
So P would be the midpoint of the slanted edge OB and joined to A?

Just making sure the diagram is correct.

#### 1729

##### Active Member
So P would be the midpoint of the slanted edge OB and joined to A?

Just making sure the diagram is correct.
P belongs to the slanted edge OB but it's not the midpoint. AP is just perpendicular to OB. If the triangle was equilateral or if AB = OA then it would be a perpendicular bisector but it isn't.

#### Andy005

##### Member
Can you show me how you would do it?

I ended up with really ugly equations and got an answer of cos114˚44', now I'm really not sure on what to do.

Thanks

#### Drongoski

##### Well-Known Member
OK - sorry don't know how to upload diagram; so stoopid!

Following 1729: let square base be labelled anticlockwise ABCD (with side AB nearest to you) and the apex E. Now consider the 2 adjacent faces: triangles AEB and AED. Let P be the common foot of the perpendiculars from B to AE and from D to AE. Then consider the triangle PBD. The required angle between the 2 faces is angle DPB of the isosceles triangle DPB.

$\bg_white \angle EAB = \theta \implies cos \theta = \frac {1}{\sqrt 6} \implies sin^2 \theta = 1 - cos^2 \theta = \frac {5}{6}\\ \\ AB = w \implies DB = \sqrt 2 w and BP = DP = wsin \theta \\ \\ \therefore by the cosine rule: cos \angle DPB = \frac {DP^2 + BP^2 - DB^2}{2 \times DP \times BP}\\ \\ = \frac {2w^2 sin^2 \theta - 2w^2}{2w^2 sin^2 \theta} = 1 - \frac {1}{sin^2 \theta} = 1 - \frac {6}{5} = -\frac {1}{5}$

That's it.

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