of the tower?
2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches
the ground by making 30° angle with the ground. The distance between the foot of the
tree and the top of the tree on the ground is 6m. Find the height of the tree before falling
down.
Answers
Answer:
Hey there !!
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
this is one of the example now I can do by urself
.
Step-by-step explanation: