diff --git a/00.brainstorming-pedagogical-teams/45.synthesis-structuring/instructions-for-levels/30.beyond/annex.fr.md b/00.brainstorming-pedagogical-teams/45.synthesis-structuring/instructions-for-levels/30.beyond/annex.fr.md
index b53d46f8c..ee80a3652 100644
--- a/00.brainstorming-pedagogical-teams/45.synthesis-structuring/instructions-for-levels/30.beyond/annex.fr.md
+++ b/00.brainstorming-pedagogical-teams/45.synthesis-structuring/instructions-for-levels/30.beyond/annex.fr.md
@@ -183,7 +183,8 @@ Sont proposées les catégories suivantes, mais à débattre, toutes les idées
 ! * The first spherical refracting surface $`DS1`$ encountered by the light has
 ! the follwing characteristics :<br>
 ! $`\overline{S_1C_1}=+|R|=+5\;cm`$ , $`n_{ini}=1`$ and $`n_{fin}=1.5`$
-! * The second spherical refracting surface $DS2$ encountered by the light has the follwing characteristics :<br>
+! * The second spherical refracting surface $DS2$ encountered by the light 
+! has the follwing characteristics :<br>
 ! $`\overline{S_2C_2}=-|R|=-5\;cm`$ , $`n_{ini}=1.5`$ and $`n_{fin}=1`$
 !
 ! * Algebraic distance between $DS1$ and $DS2$ is : $`\overline{S_1S_2}=+10\;cm`$
@@ -451,25 +452,25 @@ Sont proposées les catégories suivantes, mais à débattre, toutes les idées
 !<summary>
 ! What is the apparent magnification of the cathedral ?
 !</summary>
-! 
+! <br>
 ! * "apparent magnification" = "angular magnification" = "magnifying power".
 !
 ! * As calculated previously, standing 400 metres from the cathedral, the 90 m heigh
 ! cathedral sustends the apparent angles of $`\alpha=arctan\left(\dfrac{90}{400}\right)=0.221\;rad=12.7°`$ 
 ! at your eye.
-!
+! <br>
 ! * The image of the cathedral is 1.7 cm heigth and is located between the lens 
 ! (from its vertex $`S2`$) and your eyes and at 2.5cm from the lens. If your eye is
 ! 20cm away from the lens, so the distance eye-image is 17.5 cm (we use no algebraic values).
 ! Thus the image of the catedral subtends the apparent angle 
 ! $`\alpha'=arctan\left(\dfrac{1.7}{17.5}\right)=0.097\;rad=5.6°`$ at your eye.
-!
+! <br>
 ! * The apparent magnification $`M_A`$ of the cathedral throught the lensball for my 
 ! eye in that position is<br>
 !  $`M_A=\dfrac{\alpha'}{\alpha}=\dfrac{0.097}{0.221}=0.44`$.<br><br>
 !  Taking into account that the image is reversed, the algebraic value of the apparent
 ! magnification is $`\overline{M_A}=-0.44`$.
-!
+! <br>
 ! * You could obtained directly this algebraic value of $`M_A`$ by considering algebraic
 ! lengthes and angles values in the calculations :<br><br> 
 ! $`\overline{M_A}=\dfrac{\overline{\alpha'}}{\overline{\alpha}}`$