Question

Find the height of the tree which is broken at some point by wind and the broken part makes 30° with the ground at a distance of 10√3m from root.

solve by process ​

Answer 1

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Height of the tree = AB + AC

Trigonometric ratio which involves AB, BC and ∠C is tan θ, where AB can be measured.

Trigonometric ratio which involves AB, AC and ∠C is sin θ, where AC can be measured.

Distance between the foot of the tree to the point where the top touches the ground = BC = 8 m

In triangle ABC,

tan C = AB / BC

tan 30° = AB / 8

1/√3 = AB / 8

AB = 8 / √3

sin C = AB / AC

sin 30° = (8/√3) / AC

1/2 = 8/√3 × 1 / AC

AC = 8/√3 × 2

AC = 16 / √3

Height of tree = AB + AC

= 8/√3 + 16/√3

= 24/√3

= 24 × √3 / √3 × √3. [On rationalizing ]

= (24√3) / 3

= 8√3

So, the height of tree is 8√3 meters.

Answer 2

Answer:

In triangle ABC,

tan C = AB / BC.

tan 30° = AB / 8.

1/√3 = AB / 8

• AB = 8 / √3.

sin C = AB / AC.

sin 30° = (8/√3) / AC.

1/2 = 8/√3 × 1 / AC.

AC = 8/√3 × 2.

• AC = 16 / √3.

Height of tree = AB + AC.

= 8/√3 + 16/√3.

= 24/√3.

= 24 × √3 / √3 × √3.

= (24√3) / 3.

• = 8√3.

Hence, the height of tree is 8√3 meters.