well, i thought it was discussed here not long ago, true. If you scan the internet, it may help you a lot.

TryTHIS, it may help you.

www.cut-the-knot.org/pythagoras/cos36.shtml

It follows that every angle of the regular pentagon equals 108°. Inscribed CAD is half of the central angle of 360°/5, i.e.

CAD = 36°.

By symmetry BAC = DAE implying that this two are also 36°. (Note in passing that we just proved that angle 108° is trisectable.) Other angles designated in the diagram are also easily calculated.

As far as linear segments are concerned, we may observe that triangles ABP, ABE, AEP are isosceles, in particular,

AB = BP,

AB = AE,

AP = EP.

Triangles ABE and AEP are also similar, in particular

BE / AB = AE / EP, or

BE·EP = ABē, i.e.

(BP + EP)·EP = ABē, and lastly,

(AB + EP)·EP = ABē.

For the ratio x = AB/EP we have the equation

x + 1 = xē,

with one positive solution x = φ, the golden ratio. (This calculations were in fact performed to justify a construction of regular pentagon. We return to them here in order to facilitate the references.)

In ΔAEP AE = AB and EP is one of the sides such thatAE/EP = φ.Drop a perpendicular from P to AE to obtain two right triangles. Then say

cos(AEP) = (AE/2)/EP = (AE/EP)/2 = φ/2.

But AEP = 36° and we get the desired result.

Using cos(36°) = (1 + √5)/4 we can find

cos(18°) = √2(√5 + 5)/4

from cos(2α) = 2cosē(α) - 1 and then

sin(18°) = √2(3 - √5 )/4.

from cosē(α) + sinē(α) = 1. Now, it may be hard to believe but this expression simplifies to

<A name=sin18>sin(18°) = (√5 - 1)/4,

which is immediately verified by squaring the two expressions. Also

sin(36°) = √2(5 - √5 )/4

from sin(2α) = 2sin(α)cos(α).

We can easily find cos(72°) from cos(2α) = 2cosē(α) - 1:

cos(72°)= 2cosē(36°) - 1= 2[(√5 + 1) / 4]ē - 1= (6 + 2√5) / 8 - 1= (3 + √5) / 4 - 1= (√5 - 1) / 4.

This is of course equal to sin(18°) as might have been expected from the general formula, sin(α) = cos(90° - α).

Also,

sin(72°) = √2(5 + √5 )/4.

One Solution

The sublime triangle is an isosceles triangle with base angles and the angle opposite the

base is . Let triangle ABC be a sublime triangle with base BC, and construct the angle bisector of angle C produced to its intersection with AB at D. Let DB = b and BC = a. Then triangle CDB is similar to triangle ABC and CD = a, AD = a, AB = a + b, AC = a + b. The similarity allows the equation

and the ratio a/b is the golden ratio

the solution to the equation

or

Angle DCB is and using the law of cosines we can write

Angle CDA is and using the law of cosines we get

Therefore