Question

একটি গাছ ঝড়ে এমনভাবে ভেঙ্গে গেল যে তার ভাঙ্গা অংশ দন্ডায়মান অংশের
সাথে 30° কোণ উৎপন্ন করে এবং গাছের গােড়া থেকে 20 মিটার দূরে মাটি স্পর্শ করে
ক) তথ্যের ভিত্তিতে চিত্র অঙ্কন করে বর্ণনা দাও।
খ) গাছটির সম্পূর্ণ দৈর্ঘ্য নির্ণয় কর।
গ) যদি গাছটি ভেঙ্গে এর শীর্ষবিন্দু ভূমিতে 45° কোণ উৎপন্ন করে তবে গাছটি
কত উচুতে ভেঙ্গে গিয়েছিল তা নির্ণয় করাে ।​

Answer 1

Answer:

?kya h ye?????? answer lena h to hindi mudim

Answer 2

Given:

A tree was broken by a storm in such a way that its broken part generates an angle of 30° and touches the ground 20 m away from the trunk of the tree

To find:

A) Draw and describe based on the information.

B) Determine the full length of the tree.

C) If the tree breaks and its apex forms an angle of 45° to the ground, then determine how high the tree  was broken.

Solution:

Case (A):- Drawing and describing the given situation:

Referring to the first figure case (A), attached below, let's assume

"AB" → standing part of the tree

"A" → the tree is broken at point A

"AC" → broken part of the tree

"∠ACB = 30°" → the angle made by the broken part with the ground

"BC = 20 m" → the distance between the apex of the broken part of the tree and the trunk of the tree

Case (B):- Finding the full length of the tree:

Considering ΔABC, we have

$tan \theta = \frac{AB}{BC}$

substituting θ = 30° and BC = 20 m

$\implies tan \:30\° = \frac{AB}{20}$

$\implies \frac{1}{\sqrt{3} } = \frac{AB}{20}$

$\implies AB = \frac{20}{\sqrt{3} }$

$\implies \bold{AB = 11.54\:m }$

Similarly,

Considering ΔABC, we have

$cos \:\theta = \frac{BC}{AC}$

substituting θ = 30° and BC = 20 m

$\implies cos \:30\° = \frac{20}{AC}$

$\implies \frac{\sqrt{3}}{ 2} = \frac{20}{AC}$

$\implies AC = \frac{20\times 2}{\sqrt{3} }$

$\implies \bold{AC = 23.09\:m }$

∴ The length of the tree = AB + AC = 11.54 + 23.09 = 34.63 m

Thus, the full length of the tree is → 34.63 m.

Case (C):- Finding the height at which the tree was broken:

Referring to the second figure Case (C), let's assume,

"AB" → standing part of the tree

"A" → the tree is broken at point A

"AC" → broken part of the tree

"∠ACB = 45°" → the angle made by the broken part with the ground

"BC = 20 m" → the distance between the apex of the broken part of the tree and the trunk of the tree

Considering ΔABC, we have

$tan \theta = \frac{AB}{BC}$

substituting θ = 45° and BC = 20 m

$\implies tan \:45\° = \frac{AB}{20}$

$\implies 1 = \frac{AB}{20}$

$\implies \bold{AB = 20\:m }$

Thus, if the tree breaks and its apex forms an angle of 45° to the ground, then the height at which the tree was broken is → 20 m.

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What is the length of the broken part?

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