Question

the angle of elevation of the top of a tree from a point a on the ground is 60° on walking 20 metres away from its base to a point B the angle of elevation changes to 30° find the height of the tree​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that CD be the tree of height 'h' metres.

Let A be the point on the ground such that angle of elevation of the top of tree is 60°.

Now, He walks 20 metres away from the A to the point B such that angle of elevation of the top of the tree is 30°.

Let assume that AC = x metres.

Now, In right-angle triangle ACD

$\rm \: tan60\degree \: = \: \dfrac{CD}{AC}$

$\rm \: \sqrt{3} \: = \: \dfrac{h}{x}$

$\rm\implies \:h = \sqrt{3}x - - - - (1)$

Now, In right-angle triangle BCD

$\rm \: tan30\degree \: = \: \dfrac{CD}{BC}$

$\rm \: \dfrac{1}{ \sqrt{3} } = \dfrac{h}{20 + x}$

$\rm \: \dfrac{1}{ \sqrt{3} } = \dfrac{ \sqrt{3} x}{20 + x} \: \: \: \{using \: (1) \: \}$

$\rm \: 3x = 20 + x$

$\rm \: 3x - x = 20$

$\rm \: 2x = 20$

$\rm\implies \:x = 10 - - - - (2)$

On substituting the value of x in equation (1), we get

$\rm\implies \:\rm \: h = 10 \sqrt{3} = 10 \times 1.732 = 17.32 \: metres$

Hence,

Height of the tree = 17.32 metres.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Given :-

The angle of elevation of the top of a tree from a point a on the ground is 60° on walking 20 metres away from its base to a point B the angle of elevation changes to 30

To Find :-

Height of tree

Solution :-

Let us assume

height = 'h' metre

From the figure we concluded

DC = h = Height

∠B = 60°

∠A = 30°

BC = a

Now,

We know that

cotθ = Base/Perp

In ΔBCD

BC/CD = cot(60°)

• cot(60°) = 1/√3

BC/CD = 1/√3

BC/h = 1/√3

h = √3BC

h/√3 = BC (i)

In ΔACD

AC/CD = cot(30°)

• cot(30°) = √3

AC/CD = √3

AC/h = √3

AC = √3h

AC/h = √3 (ii)

Now

AC = AB + BC

AC - BC = AB

√3h - h/√3 = 20

Taking h as common

h(√3 - 1/√3) = 20

h(√3 × √3 - 1/√3) = 20

h[(√3)² - 1/√3] = 20

h[3 - 1/√3] = 20

h[2/√3] = 20

2h/√3 = 20

h = 20 × √3/2

h = 10 × √3

h = 10√3 m