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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: /0 on August 03, 2008, 07:19:47 pm

Title: Angle between adjacent planes
Post by: /0 on August 03, 2008, 07:19:47 pm: A right square pyramid, vertex O, stands on a square base ABCD. The height is 15 cm and base side length 10 cm. Find the magnitude of the angle between two adjacent sloping faces.

Please do not use coordinate geometry or vectors.

Thanks
Title: Re: Angle between adjacent planes
Post by: Mao on August 05, 2008, 03:44:35 pm: let the centre of the square base be X, and the midpoint of the side AB be Y.
let Z be a point on AO such that BZ is perpendicular to AO

Firstly, we need to find the angle OAB $\alpha$ :

$\tan(\alpha)=\frac{OY}{AY}=\frac{\sqrt{OX^2+XY^2}}{AY}=\frac{\sqrt{15^2+5^2}}{5}=\sqrt{10}$

$\alpha=\tan^{-1}(\sqrt{10})$

next, the magnitude of BZ:

$\sin(\alpha)=\frac{BZ}{AB}$

if we constuct a right-angled triangle with the angle $\alpha$ , opposite side of $\sqrt{10}$ and adjacent side of 1,

it can be shown that $\sin(\alpha)=\sqrt{\frac{10}{11}}$

$\sqrt{\frac{10}{11}}=\frac{BZ}{10}$

$BZ=10\sqrt{\frac{10}{11}}$

finally, the triangle XZB will be a right-angled triangle. let the angle XZB be $\beta$

$\sin(\beta)=\frac{XB}{BZ}=\frac{5\sqrt{2}}{10\sqrt{\frac{10}{11}}}=\frac{1}{2}\sqrt{\frac{11}{5}}$

hence, the angle between two sloping faces will be $2\beta$

$2\beta = 2\sin^{-1}\left(\frac{1}{2}\sqrt{\frac{11}{5}}\right)\approx 95.7^o$

is that... even remotely right? :P
Title: Re: Angle between adjacent planes
Post by: /0 on August 05, 2008, 11:08:22 pm: Thankyou mao, that is correct ! :)