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THe diagram shows triangle ABC where points A , B , C and P are coplanar . BC=BP=AP and angle BAC =angle ABP = angle PBC = pi/5 rad
Prove that triangle ABC and BPC are similar triangles . Hence deduce that
(1) BC^2=CP.CA
(2) $\displaystyle \cos (\frac{\pi}{5})=\frac{1}{4}(\sqrt{5}+1)$
I am only stucked with (2) , i am ok with the rest . THanks .
if your diagram is right, then BC = BP is wrong.
why is it soQuote:
Originally Posted by ukorov
oh alright not wrong actually, but could have been more detailed and more accurate. (Giggle)Quote:
Originally Posted by thereddevils
Hello thereddevils(2) Suppose $\displaystyle BP=BC=AP = d$. Then in $\displaystyle \triangle BPC$Quote:
Originally Posted by thereddevils
$\displaystyle \angle BCP = \tfrac{2\pi}{5}$So, using the result in (1):
$\displaystyle \Rightarrow PC = 2d\cos(\angle BCP) = 2d\cos(\tfrac{2\pi}{5})$
$\displaystyle d^2 = 2d\cos(\tfrac{2\pi}{5})\Big(d+2d\cos(\tfrac{2\pi}{ 5})\Big)$Solving the quadratic, taking the positive root gives
$\displaystyle \Rightarrow 1 = 2\cos(\tfrac{2\pi}{5})+4\cos^2(\tfrac{2\pi}{5})$
$\displaystyle \cos(\tfrac{2\pi}{5})=\frac{\sqrt5-1}{4}$Grandad
$\displaystyle \Rightarrow 2\cos^2(\tfrac{\pi}{5})-1 = \frac{\sqrt5-1}{4}$
$\displaystyle \Rightarrow \cos^2(\tfrac{\pi}{5})= \frac{\sqrt5+3}{8}$$\displaystyle =\frac{5+2\sqrt5+1}{16}$$\displaystyle \Rightarrow \cos(\tfrac{\pi}{5}) = \frac{\sqrt5+1}{4}$
$\displaystyle =\frac{(\sqrt5+1)^2}{16}$
why $\displaystyle \cos(\tfrac{2\pi}{5}) = 2\cos^2(\tfrac{\pi}{5})-1$ ??Quote:
Originally Posted by Grandad
Hello ukorovBecause $\displaystyle \cos2\theta = 2\cos^2\theta -1$.Quote:
Originally Posted by ukorov
Grandad
