A tree break due to storm and touched at ground and elevation is 30m the distance between the tree to touches the p of the ground is 8m and the distance to top of the tree to touches the ground is 10 find the total height of the tree
Answers
Given:
The broken part bends so that the top of the tree touches the ground making an angle 30 °.
The distance between the foot of the tree to the point where the top touches the ground is 8 m.
Original height of tree that is ABC
Let we consider that the tree bend by strom from the point B and angle formed by the top of the tree is 30°.
According to the diagram
Height of tree(A`BC) = AB + BC
Imagine A`BC = AC
A` is a top of tree which will be on the land form an angle but after bending height of tree remain same.
\bold{\underline{\underline{In \: \triangle{ABC} \: \colon}}}
In△ABC:
\bold{ tan\: 30^{\circ} \: = \: \dfrac{BC}{AC}}tan30
∘ =
AC
BC
\bold{ \dfrac{1}{\sqrt{3}} \: = \: \dfrac{BC}{8} }
3
1
=
8
BC
\bold{ BC \: = \: \dfrac{8}{\sqrt{3}}}BC=
3
8
Rationalising the denominator and numerator
\bold{ BC \: = \: \dfrac{8}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}}BC=
3
8
×
3
3
\bold{ BC \: = \: \dfrac{8\sqrt{3}}{(\sqrt{3})^{2}}}BC=
(
3
)
2
8
3
\bold{ BC \: = \: \dfrac{8\sqrt{3}}{3}}BC=
3
8
3
So,
\bold{ BC \: = \: \dfrac{8\sqrt{3}}{3}\: m }BC=
3
8
3
m
Now in same ∆ ABC
\bold{cos\: 30^{\circ} \: = \: \dfrac{AC}{AB}}cos30
∘
=
AB
AC
\bold{ \dfrac{\sqrt{3}}{2} \: = \: \dfrac{8}{AB} }
2
3
=
AB
8
\bold{ AB \: = \: \dfrac{8\times 2}{\sqrt{3}} }AB=
3
8×2
\bold{ AB \: = \: \dfrac{16}{\sqrt{3}} }AB=
3
16
After rationalising
\bold{ AB \: = \: \dfrac{16\sqrt{3}}{3} }AB=
3
16
3
\bold{AB \: = \: \dfrac{16\sqrt{3}}{3}}\: mAB=
3
16
3
m
Now we have ,
A`BC = AB + BC
\bold{A`BC \: = \: \dfrac{16\sqrt{3}}{3} \: + \: \dfrac{8\sqrt{3}}{3}}A‘BC=
3
16
3
+
3
8
3
Take LCM
\bold{A`BC \: = \: \dfrac{16\sqrt{3} \: + \: 8\sqrt{3}}{3}}A‘BC=
3
16
3
+8
3
\bold{A`BC \: = \: \dfrac{24\sqrt{3}}{3}}A‘BC=
3
24
3
\bold{A`BC \: = \: \dfrac{\cancel{24}\sqrt{3}}{3}}A‘BC=
3
24
3
\bold{A`BC \: = \:8\sqrt{3}\: m}A‘BC=8
3
m
\bold{\large{\underline{Answer}}}
Answer
So , original height of tree is \bold{\boxed{\boxed{8\sqrt{3}\: m}}}
8
3
m
Answer:
Given:
The broken part bends so that the top of the tree touches the ground making an angle 30 °.
\bold{\angle{BAC}\: = \: 30^{\circ}}∠BAC=30
∘
The distance between the foot of the tree to the point where the top touches the ground is 8 m.
\bold{\large{\underline{To\: find\: \colon}}}
Tofind:
Original height of tree that is ABC
< /p > < p > \bold{\large{\underline{Solution\: \colon}}} </p><p>
Solution:
Let we consider that the tree bend by strom from the point B and angle formed by the top of the tree is 30°.
According to the diagram
Height of tree(A`BC) = AB + BC
Imagine A`BC = AC
A` is a top of tree which will be on the land form an angle but after bending height of tree remain same.
\bold{\underline{\underline{In \: \triangle{ABC} \: \colon}}}
In△ABC:
\bold{ tan\: 30^{\circ} \: = \: \dfrac{BC}{AC}}tan30
∘ =
AC
BC
\bold{ \dfrac{1}{\sqrt{3}} \: = \: \dfrac{BC}{8} }
3
1
=
8
BC
\bold{ BC \: = \: \dfrac{8}{\sqrt{3}}}BC=
3
8
Rationalising the denominator and numerator
\bold{ BC \: = \: \dfrac{8}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}}BC=
3
8
×
3
3
\bold{ BC \: = \: \dfrac{8\sqrt{3}}{(\sqrt{3})^{2}}}BC=
(
3
)
2
8
3
\bold{ BC \: = \: \dfrac{8\sqrt{3}}{3}}BC=
3
8
3
So,
\bold{ BC \: = \: \dfrac{8\sqrt{3}}{3}\: m }BC=
3
8
3
m
Now in same ∆ ABC
\bold{cos\: 30^{\circ} \: = \: \dfrac{AC}{AB}}cos30
∘
=
AB
AC
\bold{ \dfrac{\sqrt{3}}{2} \: = \: \dfrac{8}{AB} }
2
3
=
AB
8
\bold{ AB \: = \: \dfrac{8\times 2}{\sqrt{3}} }AB=
3
8×2
\bold{ AB \: = \: \dfrac{16}{\sqrt{3}} }AB=
3
16
After rationalising
\bold{ AB \: = \: \dfrac{16\sqrt{3}}{3} }AB=
3
16
3
\bold{AB \: = \: \dfrac{16\sqrt{3}}{3}}\: mAB=
3
16
3
m
Now we have ,
A`BC = AB + BC
\bold{A`BC \: = \: \dfrac{16\sqrt{3}}{3} \: + \: \dfrac{8\sqrt{3}}{3}}A‘BC=
3
16
3
+
3
8
3
Take LCM
\bold{A`BC \: = \: \dfrac{16\sqrt{3} \: + \: 8\sqrt{3}}{3}}A‘BC=
3
16
3
+8
3
\bold{A`BC \: = \: \dfrac{24\sqrt{3}}{3}}A‘BC=
3
24
3
\bold{A`BC \: = \: \dfrac{\cancel{24}\sqrt{3}}{3}}A‘BC=
3
24
3
\bold{A`BC \: = \:8\sqrt{3}\: m}A‘BC=8
3
m
\bold{\large{\underline{Answer}}}
Answer
So , original height of tree is \bold{\boxed{\boxed{8\sqrt{3}\: m}}}
8
3
m