Question

a tree breaks due to storm and its upper end touches the ground and a make an angle of a measure 30 degree with the ground if the length of upper part of the tree is 10√3 then find the total height of tree. as when the distance between foot of the tree and where it touches the ground​

Answer 1

Answer:

• Total height of tree is 15√3.
• Length of distance between foot of tree and top of tree where it is touching ground is 15 .

Step-by-step explanation:

Given :-

• Length of upper part of tree is 10√3.
• Angle of elevation is 30°.

To find :-

• Total height of tree.
• Distance between foot of tree and top of tree where it touching ground.

Solution :-

Let, Total or we can say original height of tree be h.

Length of lower part of tree be a.

And Length of distance between foot of tree and top of tree where it is touching ground be b.

So,

$\sf \implies sin \: 30\degree = \dfrac{a}{10\sqrt{3}}$

• Sin 30° = 1/2.

$\sf \implies \dfrac{1}{2} = \dfrac{a}{10 \sqrt{3}}$

• Cross multiple.

$\sf \implies 10 \sqrt{3} = 2 a$

$\sf \implies \dfrac{10 \sqrt{3}}{2} = a$

$\sf \implies \bold{a = 5\sqrt{3}}$

a is length of Lower part.

Length of lower part of tree is 5√3.

Total height of tree = Length of upper part of tree + Length of lower part of tree

$\sf \implies h = 10 \sqrt{3} + 5 \sqrt{3}$

$\sf \implies \bold{h = 15 \sqrt{3}}$

• Or, 25.98 [15 × 1.7330 = 25.98]

Thus,

Total height of tree is 15√3 or 25.98 .

Now,

By Pythagoras theorem :

• Base² = Hypotenuse² - Perpendicular²

Base = b

Hypotenuse = 10√3

Perpendicular = 5√3

Put all values :

$\sf \implies b^{2} = (10\sqrt{3})^{2} - (5 \sqrt{3})^{2}$

$\sf \implies b^{2} = 300 - 75$

$\sf \implies b^{2} = 225$

$\sf \implies b = \sqrt{225}$

$\sf \implies \bold{b = 15}$

We take b be distance between foot and top of tree.

Thus,

Length of distance between foot of tree and top of tree where it is touching ground is 15 .

Answer 2

$\: \: \color{blue}\underline{ \large\rm{Solution :- }}$

Let the broken part of tree be AC

It is given that,

Distance between foot of tree B and point C = 8 cm

so, BC = 8 cm

Also, broken parts of tree makes an angle 30° with ground

so , ∠ C = 30°

we need to find height of the tree

$\\ \: \: \: \color{grey} \sf \: Height \: of \: tree = \: height \: of \: broken \: part \: + \: height \: of \: remaining \: tree$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies\: \sf \: Height \: of \: tree = AB + AC$

Since,

The tree was vertical to the ground

so , ∠ ABC = 90°

In right angle triangle ABC ,

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \: \implies \sf \: cos \: C \: = \frac{side \: adjacent \: to \: angle \:C}{hypotenuse}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \: \implies \sf \: cos \: C \: = \frac{BC}{AC}$

$\\ \: \: \: \: \: \: \: : \: \implies \sf \: cos \: 30\: = \frac{8}{AC}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: : \: \sf\implies \sf \: \frac{ \sqrt{3} }{2} = \frac{8}{AC}$

$\\ \: \: \: \: \: \: \:\: : \: \sf\implies \sf AC\: = \frac{8 \times 2}{ \sqrt{3} }$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: : \: \frak{\implies \pink{ \boxed{ { \frak{AC\: = \frac{8 \times 2}{ \sqrt{3} }}} }}}$

In right angle triangle ABC ,

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \: \implies \sf \: cos \: C \: = \frac{side \: adjacent \: to \: angle \:C}{hypotenuse}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \: \implies \sf \: cos \: 30 \degree \: = \frac{AB}{AC}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: : \: \sf\implies \sf \: \frac{ 1}{2} = \frac{AB}{ \frac{16}{ \sqrt{3} } }$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: : \: \sf\implies \sf \: \frac{ 1}{2} = \frac{ \sqrt{3} }{16} \times AB$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \:\: : \: \pink{ \boxed{\implies \frak {AB= \frac{ 16 }{2 \sqrt{3} } = \frac{8}{ \sqrt{3} } }}}$

$\\ \: \: \: \: \: : : \sf \: so. \: \: \: \sf \: Height \: of \: tree = AB + AC$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \sf \: \frac{16}{ \sqrt{3} } + \frac{8}{ \sqrt{3} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \pink{ \boxed{\frak{\frac{24}{ \sqrt{3} } }} }$

Multipling √3 in numerator and denominator,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf \: \frac{24}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf \: {24} \times \frac{ \sqrt{3} }{ {3} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: = \pink{ \boxed{\frak{8 \sqrt{3} }} }$

Hence , the height of tree is $\rm {8 \sqrt{3}$.