Let the broken portion of tree AA’ be x. Hence A’C = A’G = x
From the figure, total height of the tree = x + y + 6
Consider triangle A’BG, tan 45° = A'B / BG
1 =
Or BG = y + 6
Consider triangle A’C’C, tan 30° = A’B/C’C
=
=
y√3 = y + 10
or y =
= *
=
Therefore y = 5(√3 + 1)
Take sin 30° to find x
sin 30° = A'C' / A'C
= y/x
or x = 2y
Height of the tree = x + y = 6
= 2 * 5(√3 + 1) + 5(√3 + 1) + 6
= 10√3 + 10 + 5√3 + 5 + 6
= 15√3 + 21 meters
Answer choice (C)
Correct Answer: 15√3 + 21
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