1. ## Constructions

Construct a triangle ABC in which BC = 7 cm, <A = 60o and altitude through A is 3.7 cm. How many such triangles are possible? Please tell me the steps of construction.

2. Originally Posted by ice_syncer
Construct a triangle ABC in which BC = 7 cm, <A = 60o and altitude through A is 3.7 cm. How many such triangles are possible? Please tell me the steps of construction.
Description:

1. Draw line BC
2. Draw a parallel to BC with the perpendicular distance of $h_A$
3. Draw the perpendicular bisector of BC
4. Draw in B (or in C) the angle of (90° - <(A)) = 30° , that means <(MBC) = 30°
5. The arm of this angle crosses the perpendicular bisector in M.
6. Draw the circle around M with radius r = MB (or MC)
7. The intersection points of the circle and the parallel to BC are the points A.

3. I've attached the construction.

4. Originally Posted by earboth
I've attached the construction.
Where's the altitude through A ?

5. Originally Posted by ice_syncer
Where's the altitude through A ?
The altitude is the distance between the two parallel lines and because the angle at b (or C respectively) is greater than 90° the altitude lies outside the area of the triangle.