Question

14. The top of a tall tree broken by wind falls to the ground at an angle of 30 ° at a distance of 6 m from
the foot of the tree. Find the original height of the tree correct to nearest meter.(V3 =1.73) ​

Answer 1

Answer:

The top of a tall tree broken by wind falls to the ground at an angle of 30 ° at a distance of 6 m from

the foot of the tree. Find the original height of the tree correct to nearest meter

Answer 2

Given:-

The top of a tall tree broken by wind falls to ground at an angle of 30°

Distance from the foot of the tree = 6m

To Find:-

The original height of the tree.

Note:-

Refer to the attachment for a better explanation.

Solution:-

From the figure we can clearly see that the top of the tree broke away in such a way that it formed a right-angle triangle.

Firstly name the triangle so formed.

Now,

In ∆ABC

$\sf{\angle ACB = 30^\circ}$

$\sf{BC = 6\:m}$

We first need to find the perpendicular,

Therefore,

$\sf{Tan30^\circ = \dfrac{Perpendicular}{Base}}$

= $\sf{\dfrac{1}{\sqrt{3}} = \dfrac{AB}{6}}$

= $\sf{AB = \dfrac{6}{\sqrt{3}}}$

Rationalizing the denominator,

$\sf{AB = \dfrac{6\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}}$

= $\sf{AB = \dfrac{6\sqrt{3}}{3}}$

= $\sf{AB = 2\sqrt{3}}$

Taking √3 as 1.73

$\sf{AB = 2\times 1.73}$

= $\sf{AB = 3.46\:m}$

Therefore length of tree from the ground to the broken part is 3.46 m = 3.5 m

$\sf{\underline{\red{So \:to \:find \:the\: original \:length \:of \:the \:tree \:we \:need \:to \:find \:the\: hypotenuse\: of\: the\: \Delta{ABC}}}}$

According to Pythagoras Theorem,

$\sf{(AC)^2 = (AB)^2 + (BC)^2}$

= $\sf{AC = \sqrt{(3.5)^2 + (6)^2}}$

= $\sf{AC = \sqrt{12.25 + 36}}$

= $\sf{AC = \sqrt{48.25}}$

= $\sf{AC = 6.9\:m}$

Original Length of tree = AC + AB

= $\sf{Original\:Length\:of\:Tree = 3.5 + 6.9 = 10.4\:m}$

$\sf{\therefore The\:original\:length\:of\:the\:tree\:is\:10.4\:m}$

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Note:-

This table is very much important for solving sums related to trigonometric ratios.

$\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}$

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