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The upper part of a tree, broken over by the wind, makes an angle of 45° with the ground; and the distance from the root to the point where the top of the tree touches the ground, is 15 m. What was the height of the tree before it was broken?

Heights & Distances

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Answer

Let A be the point from where tree breaks and C be the point where above part of tree touches the ground.

The upper part of a tree, broken over by the wind, makes an angle of 45° with the ground; and the distance from the root to the point where the top of the tree touches the ground, is 15 m. What was the height of the tree before it was broken? Heights and Distances, Concise Mathematics Solutions ICSE Class 10.

From figure,

In △ABC,

tan 45°=PerpendicularBase⇒1=ABBC⇒AB=BC=15 meters.\text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{AB}{BC} \\[1em] \Rightarrow AB = BC = 15 \text{ meters}.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ AC2 = 152 + 152

⇒ AC2 = 225 + 225

⇒ AC2 = 450

⇒ AC = 450=152\sqrt{450} = 15\sqrt{2} meters.

Height of tree = AB + AC = 15 + 15215\sqrt{2}

= 15 + 21.21

= 36.21 meters.

Hence, height of tree before it was broken = 36.21 meters.

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