Question

28. The upper part of a tree is broken by the wind and makes an angle of 30" with the ground. The
distance from the foot of the tree to the point where the top touches the ground is 5 m The height of
the tree is
(a) 10V33 m
(b) 5 V33 m
C) (d) 3/5 m​

Answer 1

✬ Tree = 5√3 m ✬

Step-by-step explanation:

Given:

• Broken part of tree makes an angle of 30° with ground.
• Broken part touches the ground at the distance of 5 m from foot of tree.

To Find:

• What is the height of tree ?

Solution: Let BD be the tree and it broke from A.

Here we have

• BD = tree

• BD = AB + AD

• AD = AC (broken part of tree which fell on ground)

• BC = distance between broken part and foot of tree.

In ∆ABC , using tanθ

• AB {perpendicular}
• BC {base}
• AC {hypotenuse}

➮ tanθ = perpendicular/base

➮ tan30° = AB/BC

➮ 1/√3 = h/5

➮ 5 = √3h

➮ 5√3 = h

➮ 5/√3 = AB

For same ∆ , let's use cosθ

➮ cosθ = base/hypotenuse

➮ cos30° = BC/AC

➮ √3/2 = 5/AC

➮ AC√3 = 5 × 2

➮ AC = 10√3

Since , BD = AB + AD and AD = AC

∴ BD = AB + AC

➠ BD = 5/√3 + 10/√3

➠ Tree = 15/√3 or 15/√3 × √3/√3

• By rationalising the denominator.

➠ Tree = 15√3/3 = 5√3

Hence, the height of tree was 5√3 m.

Answer 2

Given :

• The upper part of a tree is broken by the wind and makes an angle of 30" with the ground.

• The distance from the foot of the tree to the point where the top touches the ground is 5 m

To find :

• The height of the tree is

Solution :

From the figure :

Cos 30° = AB/BC

$\sf : \implies \: \: \: \: \: \: \dfrac{ \sqrt{3} }{2} = \dfrac{5}{x} \\ \\ \\ \sf : \implies \: \: \: \: \: \: x \sqrt{3 } = 5 \times 2 \\ \\ \\ \sf : \implies \: \: \: \: \: \: x = \frac{10}{ \sqrt{3} }$

Again , tan 30° = AC / AB

$\sf : \implies \: \: \: \: \: \: \dfrac{1}{ \sqrt{3} } = \dfrac{y}{5} \\ \\ \\ \sf : \implies \: \: \: \: \: \: y \sqrt{3} = 5 \\ \\ \\ \sf : \implies \: \: \: \: \: \: y = \frac{5}{ \sqrt{3} }$

The height h of the tree is given by :

h = x + y

$\sf : \implies \: \: \: \: \: \: h = \frac{10}{ \sqrt{3} } + \dfrac{5}{ \sqrt{3} } \\ \\ \\ \sf : \implies \: \: \: \: \: \: h = \frac{15}{ \sqrt{3} } \\ \\ \\ \sf : \implies \: \: \: \: \: \: h = \frac{15}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ \\ \sf : \implies \: \: \: \: \: \: h = \frac{15 \sqrt{3} }{3} \\ \\ \\ \sf : \implies \: \: \: \: \: \: h = 5 \sqrt{3}$