Sam and markus are standing on level ground 100m apart. A large tree is due north of markus and on a bearing of 65degrees from sam. The top of the tree appears at an angle of elevation of 25degrees to sam and 15degrees to markus. Find the height of the tree!

Answer 1

Draw a sketch such that the tree (T) is due north of Markus (M), and Sam (S) is due (approx.) NW of M, such that angle STM is 65°, since the bearing of T at S is 65°.

We do not know the angles S (α) nor M (β), nor the distances ST (x1) nor MT(x2). However, we do know the distance SM = 100m.

see
http://img600.imageshack.us/img600/7954/1289559223.jpg

Let the height of the tree be h, express x1 and x2 in terms of h.

Use the sine rule to relate α, β and the known angle 65°.
This way, there will be 2 equations relating sin(α), sin(β) and h.

The third relation is given by the fact that α+β+65°=180° (angles of a triangle).

Solve by trial and error for α=80°, β=34° and h=29m approximately.

Also, for information (approx.):
x1=108.8m and x2=62.5m.

Redo and check my work.

Answer 2

9+6