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A tree is broken at a height of 10 m above the ground. The broken part touches the ground and makes an angle of 30° with the horizontal. Hence find the height of the tree ​



Let PQ be the original height of tree.Let O be the point from where it is broken, such that QO = 10 mSuppose the broken part OP that bends at O, touches the ground at O'.Now, ∠OO'Q = 30°In ∆OQO'sin 30° = OQOO'⇒12 = 10OO'⇒OO' = 20So, total height of tree = 10 + 20 = 30 m

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This is a text book question. Refer to the exercise
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Not exactly. But similar one
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Let the height of the triangle ( the tree part thats still standing ) be 10m
angle of elevation= /_ 30
using sin 30= 1/2
(sin 0= opp/hypo)
sin 30= 10/hypo
1/2=10/hypo
cross multiplying,
hypo = 20 m
Hypo is nothing but the broken part of the tree, so adding the height of triangle and the hypotenuse,
we get the complete height,
10+20= 30 m ans.
hope it helped ( also that its right ;) )
regards

 
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