# Help Géo !!!

• Feb 4th 2010, 10:06 AM
Perelman
Help Géo !!!
Hii !

I need Help For This Exercice :

Either $ABC$ a triangle and $D$ point belonging to $[BC]$ such as :

$\widehat{DAC} = 90$ , $\widehat{BAD} = 30$ , $BD=AC=1$ .

Find / Calculate $CD$ .
• Feb 4th 2010, 10:59 PM
earboth
Quote:

Originally Posted by Perelman
Hii !

I need Help For This Exercice :

Either $ABC$ a triangle and $\bold{\color{red}D}$ point <<<<<< typo?
belonging to $[BC]$ such as :

$\widehat{DAC} = 90$ , $\widehat{BAD} = 30$ , $BD=AC=1$ .

Find / Calculate $CD$ .

This is only a first step and not the final solution.

The radii have the length 1.
• Feb 5th 2010, 01:56 AM
earboth
Quote:

Originally Posted by Perelman
Hii !

I need Help For This Exercice :

Either $ABC$ a triangle and $A$ point belonging to $[BC]$ such as :

$\widehat{DAC} = 90$ , $\widehat{BAD} = 30$ , $BD=AC=1$ .

Find / Calculate $CD$ .

1. In the right triangle DCS you have:

$1+s^2 = x^2$..........[1]

2. In the triangle ABD you have:

$s \cdot \sin(30^\circ) = p~\implies~p=\tfrac12 s$

3. In the right triangle left of A you have:

$h = 1 \cdot \sin(60^\circ)~\implies~h=\tfrac12 \cdot \sqrt{3}$

4. Now use similar right triangles:

$\dfrac ph = \dfrac1{1+x}~\implies~\dfrac{\tfrac12 s}{\tfrac12 \cdot \sqrt{3}}=\dfrac1{1+x}$

which yields: $s^2 = \dfrac3{(1+x)^2}$ ..........[2]

5. Combine [1] and [2] and you have to solve for x:

$1+ \dfrac3{(1+x)^2} = x^2$

6. After moving some stuff around you'll get:

$\tfrac12 x^4 + x^3 - x -2 = 0$

7. You can show that x = -2 must be a solution of the equation, that means you can factor the LHS:

$\frac12 \cdot (x+2)(x^3-2)=0$

8. Therefore the only possible solution is $x = \sqrt[3]{2}~\approx~ 1.259921...$