Math Is Fun Forum

Identity

Member

Registered: 2007-04-18

Posts: 934

Triangle Problem

The problem is attached, the problem I'm having is I can get up to:

, but I don't know the length of CD  =[.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Triangle Problem

ΔCDB is an isosceles triangle, so CD = BC = x.

Let BD = y. Then by similar triangles, BD∕BC = BC∕BA ⇒ y∕x = x∕1 ⇒ y = x[sup]2[/sup].

So AD = 1−x[sup]2[/sup].

Applying the cosine rule to ΔABC:
BC[sup]2[/sup] = x[sup]2[/sup] = 1[sup]2[/sup] + 1[sup]2[/sup] − 2cos36°
x[sup]2[/sup] − 2 =  −2cos36°

Applying the cosine rule to ΔADC:
AD[sup]2[/sup] = (1−x[sup]2[/sup])[sup]2[/sup] = x[sup]2[/sup] + 1[sup]2[/sup] − 2xcos36°
1 − 2x[sup]2[/sup] + x[sup]4[/sup] = x[sup]2[/sup] + 1 − 2xcos36°
x[sup]3[/sup] − 3x = −2cos36°

Substituting for −2cos36° from previously,
x[sup]3[/sup] − 3x = x[sup]2[/sup] − 2
x[sup]3[/sup] − x[sup]2[/sup] − 3x + 2 = 0
(x − 2)(x[sup]2[/sup] + x − 1) = 0

But x ≠ 2 because BC must be shorter than AB and AC. Hence, x[sup]2[/sup] + x − 1 = 0
­

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Triangle Problem

Hey I just realised an easier way!

AD = DC (isosceles triangle ADC)
1−x² = x
x²+x-1=0