Question

The upper part of a tree broken by wind makes an angle ${30^\circ }$ with the ground and the distance from the foot to the point where the top of the tree meets the ground is $8$ meters. What was the height of the tree?

Answer 1

Hint:
Assume the broken part of the tree to be l m and the remaining part of the tree to be x m. Then the total height= x + l. Now use the formula of trigonometric ratio which is given as-
$ \Rightarrow \tan \theta = \dfrac{P}{B}$ where P is perpendicular and B is base. Put the values to find x. Then use $\cos \theta = \dfrac{B}{H}$ where B is the base and H is the hypotenuse of the triangle. Put the values to find l. Then put these values of x and l in the main equation of total height.

Complete step by step solution:
Let the tree be AB and it is broken from the point C so that A make angle ${30^\circ }$ with the ground.
Let the broken part be l meter.
The distance from the foot of the tree to where the top meets ground AB = $8$ meters
We have to find the height of the tree. Let the total height of the tree.
Let the point CB be x m then total height= x + l --- (i)
We have to find the value of x and l

In triangle ABC,
$ \Rightarrow \tan \theta = \dfrac{P}{B}$ where P is perpendicular and B is base.
So from the diagram we can write,
$ \Rightarrow \tan {30^ \circ } = \dfrac{{CB}}{{AB}}$
On putting the given values we get,
$ \Rightarrow \tan {30^ \circ } = \dfrac{x}{8}$
Now we know the value of trigonometric ratio $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$
On putting this value in the equation we get,
$ \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{x}{8}$
On adjusting the equation, we get,
$ \Rightarrow x = \dfrac{8}{{\sqrt 3 }}$ -- (ii)
From the diagram we can also write,
$ \Rightarrow \cos \theta = \dfrac{B}{H}$
On putting the values of base and hypotenuse, we get-
$ \Rightarrow \cos {30^ \circ } = \dfrac{{AB}}{{CA}}$
On putting the values of the sides we get,
$ \Rightarrow \cos {30^ \circ } = \dfrac{8}{l}$
Now we know that $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
On putting this value we get,
$ \Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{8}{l}$
On cross multiplication we get,
$ \Rightarrow l = \dfrac{{8 \times 2}}{{\sqrt 3 }}$
On solving we get,
$ \Rightarrow l = \dfrac{{16}}{{\sqrt 3 }}$ -- (iii)
On substituting the values of x and l from eq. (ii) and (iii) in eq. (i) we get,
Total height=$\dfrac{8}{{\sqrt 3 }} + \dfrac{{16}}{{\sqrt 3 }}$
On taking LCM we get,
Total height=$\dfrac{{8 + 16}}{{\sqrt 3 }} = \dfrac{{24}}{{\sqrt 3 }}$
On rationalizing we get,
Total height=$\dfrac{{24}}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} = \dfrac{{24\sqrt 3 }}{3} = 8\sqrt 3 $ m

Hence, the total height of the tree was $8\sqrt 3 $ m.

Note:
Here we have not used sine ratio because we know that sine ratio is given as-
$ \Rightarrow \sin {\theta ^ \circ } = \dfrac{P}{H}$
Then from the diagram when we put the values of perpendicular and hypotenuse we get,
$ \Rightarrow \sin {30^ \circ } = \dfrac{{CB}}{{CA}}$
Here we do not know the value of either CB or CA. We just assume them to be x and l meters respectively. And we have to find the value of both x and l to get the height of tree. Hence we use tan and cosine ratio.