+25 Jack and Jill are standing on level ground 100m apart. A large tree is due North of Jack and on a bearing of 065 from Jill. The top of the tree appears at an angle of elevation of 25 degrees to Jill and 15 degrees to Jack. Find the height of the tree?
+129849 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 ..........
