# CAT Practice : Trigonometry

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## Height of a tree

Q.15: A tall tree AB and a building CD are standing opposite to each other. A portion of the tree breaks off and falls on top of the building making an angle of 30°. After a while it falls again to the ground in front of the building, 4 m away from foot of the tree, making an angle of 45°. The height of the building is 6 m. Find the total height of the tree in meters before it broke.
1. 27√3 + 39
2. 12√3 + 10
3. 15√3 + 21
4. Insufficient Data

Choice C. (15√3 + 21) m

## Detailed Solution

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 = $\frac{y + 6}{BG}$
Or BG = y + 6

Consider triangle A’C’C, tan 30° = A’B/C’C
$\frac{1}{√3}$ = $\frac{y}{BG + 4}$
$\frac{1}{√3}$ = $\frac{y}{y + 6 + 4}$
y√3 = y + 10
or y = $\frac{10}{√3 - 1}$
= $\frac{10}{√3 - 1}$ * $\frac{√3 + 1}{√3 + 1}$
= $\frac{10(√3 + 1)}{2}$
Therefore y = 5(√3 + 1)

Take sin 30° to find x
sin 30° = A'C' / A'C
$\frac{1}{2}$ = 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

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