2 triangles

thereddevils · #1 $Old$ November 7th, 2009, 11:41 PM

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) $\cos (\frac{\pi}{5})=\frac{1}{4}(\sqrt{5}+1)$

I am only stucked with (2) , i am ok with the rest . THanks .

ukorov · #2 $Old$ November 8th, 2009, 01:20 AM

if your diagram is right, then BC = BP is wrong.

thereddevils · #3 $Old$ November 9th, 2009, 07:42 AM

Quote:

Originally Posted by ukorov $View Post$

if your diagram is right, then BC = BP is wrong.

why is it so

ukorov · #4 $Old$ November 9th, 2009, 12:15 PM

Quote:

Originally Posted by thereddevils $View Post$

why is it so

oh alright not wrong actually, but could have been more detailed and more accurate.

Grandad · #5 $Old$ November 10th, 2009, 03:38 AM

Hello thereddevils

Quote:

Originally Posted by thereddevils $View Post$

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) $\cos (\frac{\pi}{5})=\frac{1}{4}(\sqrt{5}+1)$

I am only stucked with (2) , i am ok with the rest . THanks .

(2) Suppose $BP=BC=AP = d$ . Then in $\triangle BPC$

$\angle BCP = \tfrac{2\pi}{5}$

$\Rightarrow PC = 2d\cos(\angle BCP) = 2d\cos(\tfrac{2\pi}{5})$

So, using the result in (1):

$d^2 = 2d\cos(\tfrac{2\pi}{5})\Big(d+2d\cos(\tfrac{2\pi}{5})\Big)$

$\Rightarrow 1 = 2\cos(\tfrac{2\pi}{5})+4\cos^2(\tfrac{2\pi}{5})$

Solving the quadratic, taking the positive root gives

$\cos(\tfrac{2\pi}{5})=\frac{\sqrt5-1}{4}$

$\Rightarrow 2\cos^2(\tfrac{\pi}{5})-1 = \frac{\sqrt5-1}{4}$

$\Rightarrow \cos^2(\tfrac{\pi}{5})= \frac{\sqrt5+3}{8}$
$=\frac{5+2\sqrt5+1}{16}$

$=\frac{(\sqrt5+1)^2}{16}$

$\Rightarrow \cos(\tfrac{\pi}{5}) = \frac{\sqrt5+1}{4}$