Question

The upper part of a tree broken by wind, falls to the ground without being detached. The top of the broken part touches the ground at an angle of 30° at a point 6 m from the foot of the free. Calculate the original height of the tree correct to two decimal places.​

Answer 1

The original height of the tree is $6\sqrt{3}$ m or 10.39 meters.

Step-by-step explanation:

We are given that the upper part of a tree broken by the wind, falls to the ground without being detached. The top of the broken part touches the ground at an angle of 30° at a point 6 m from the foot of the free.

As we can see in the figure that the length of AB and A'B will be same because A'B is just the broken part of the original tree of height AC, that means;

The height of the original tree = AB + BC

Also, the side A'C is of 6 m and the angle BA'C is of 30°.

Now, in the triangle BA'C;

 $\text{tan } 30\° = \frac{\text{Perpendicular}}{\text{Base}}$

  $\frac{1}{\sqrt{3} } =\frac{BC}{A'C}$

  $BC = \frac{6}{\sqrt{3} }$

 BC = $2\sqrt{3}$ m

Similarly, in the triangle BA'C;

   $\text{cos } 30\° = \frac{\text{Base}}{\text{Hypotenuse}}$

  $\frac{\sqrt{3}}{2 } =\frac{A'C}{A'B}$

  $A'B = \frac{6 \times 2 }{\sqrt{3} }$

 A'B = $4\sqrt{3}$ m

Hence, the original height of the tree = A'B + BC = $4\sqrt{3}$ m + $2\sqrt{3}$ m = $6\sqrt{3}$ m.