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!

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  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.


    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.

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