1. ## Help Géo !!!

Hii !

I need Help For This Exercice :

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

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

Find / Calculate $\displaystyle CD$ .

2. Originally Posted by Perelman
Hii !

I need Help For This Exercice :

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

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

Find / Calculate $\displaystyle CD$ .
This is only a first step and not the final solution.

The radii have the length 1.

3. Originally Posted by Perelman
Hii !

I need Help For This Exercice :

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

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

Find / Calculate $\displaystyle CD$ .
1. In the right triangle DCS you have:

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

2. In the triangle ABD you have:

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

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

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

4. Now use similar right triangles:

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

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

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

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

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

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

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

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

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