A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree
Answers
Step-by-step explanation:
Let AB be the original height of the tree.
Suppose it got bent at a point C and let the
part CB take the position CD, meeting the ground at D. Then,
→ AD = 8m, \angle∠ ADC = 30° and CD = CB.
→ Let AC = x metres and CD = CB = y metres.
From right ∆DAC, we have
=> \frac{AC}{AD} = tan 30 \degree = \frac{1}{ \sqrt{3} } .
AD
AC
=tan30°=
3
1
.
=> \frac{x}{8} = \frac{1}{ \sqrt{3} } .
8
x
=
3
1
.
=> x = \frac{8}{ \sqrt{3} } .
3
8
.
=> x = \frac{8}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } .
3
8
×
3
3
.
=> x = \frac{ 8 \sqrt{3} }{ 3 } .
3
8
3
.
▶ Also, from right ∆DAC, we have
=> \frac{CD}{AD} = sec 30 \degree = \frac{2}{ \sqrt{3} } .
AD
CD
=sec30°=
3
2
.
=> \frac{y}{8} = \frac{2}{ \sqrt{3} } .
8
y
=
3
2
.
=> y = \frac{16}{ \sqrt{3} } .
3
16
.
=> y = \frac{16}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } .
3
16
×
3
3
.
=> y = \frac{ 16 \sqrt{3} }{ 3 } .
3
16
3
.
\therefore AC = \frac{ 8 \sqrt{3} }{ 3 } m and CB = \frac{ 16 \sqrt{3} }{ 3 } m.∴AC=
3
8
3
mandCB=
3
16
3
m.
▶ Total height of tree = AC + BC.
= \frac{ 8 \sqrt{3} }{ 3 } + \frac{ 16 \sqrt{3} }{ 3 }=
3
8
3
+
3
16
3
= \frac{ 8 \sqrt{3} + 16 \sqrt{3} }{3} .=
3
8
3
+16
3
.
= \frac{ 24 \sqrt{3} }{3} .=
3
24
3
.
\huge \boxed{ \boxed{ \bf = 8 \sqrt{3} m. }}
=8
3
m.
✔✔ Hence, it is solved ✅✅.
____________________________________
\huge \boxed{ \boxed{ \boxed{ \mathbb{THANKS}}}}
THANKS
\huge \bf{ \#BeBrainly.}#BeBrainly.
Using given instructions, draw a figure. Let AC be the broken part of the tree. Angle C = 30°
BC = 8 m
To Find: Height of the tree, which is AB
Total height of the tree is the sum of AB and AC i.e. AB+AC
In right ΔABC,
Using Cosine and tangent angles,
cos 30° = BC/AC
We know that, cos 30° = √3/2
√3/2 = 8/AC
AC = 16/√3 …(1)
Also,
tan 30° = AB/BC
1/√3 = AB/8
AB = 8/√3 ….(2)
Therefore, total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3 = 8√3 m.