Let Jack's distance from the tree = J1 and Jill's distance from the tree = J2
Then.....letting h be the height of the tree....we have....
tan 15 = h / J1 and tan 25 = h / J2
Thus h = J1*tan 15
And tan25 = [J1*tan 15] / J2 → tan 25/tan15 = J1/J2 = about 1.74......so Jack is about 1.74 times as far from the tree as Jill......if we let Jill's distance = m ....then Jack's = 1.74m
And....using the Law of Cosines, we have :
100^2 = (m)^2 + (1.74m)^2 - 2(m)(1.74m)cos65
This is a little sticky to solve.....so I'll let WolframAlpha do the "heavy lifting".....solving for m, we have m ≈ 62.538m
So....Jill is ≈ 62.538m from the tree and Jack is ≈ 1.74(62.538)m ≈ 108.816 m from the tree
So
tan15 = h/108.816
108.816*tan15 = h = about 29.157m
tan25 = h/62.538
62.538*tan25 =h = about 29.16 m
Close enough !!!....we'll call the height of the tree 29.16 m ..........
