Question

Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30, respectively. Find the height of the poles and the distances of the point from the poles.​

Answer 1

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

Given that,

Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide.

From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively.

Let assume that AB and CD be equal poles of height h such that AC = 80 m

Now, Let assume that P be any point on the road AC such that AP = x m and PC = (80 - x) m.

Now, In right-angle triangle APB

$\rm \: tan60 \degree \: = \: \dfrac{AB}{AP}$

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

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

Now, In right-angle triangle DCP

$\rm \: tan30 \degree \: = \: \dfrac{DC}{CP}$

$\rm \: \frac{1}{ \sqrt{3} } \: = \: \dfrac{h}{80 - x}$

On substituting the value of h from equation (1), we get

$\rm \: \frac{1}{ \sqrt{3} } \: = \: \dfrac{ \sqrt{3} x}{80 - x}$

$\rm \: 3x = 80 - x$

$\rm \: 4x = 80$

$\rm\implies \:x = 20 \: m$

So, on substituting x = 20 in equation (1), we get

$\rm \: h \: = \: 20 \sqrt{3} \: m$

$\rm\implies \:\boxed{ \rm{ \:Height \: of \: pole \: = \: 20 \sqrt{3} \: m \: }}$

Now,

Distance of the first pole AB from P = x = 20 m

Distance of the second pole CD from P = 80 - 20 = 60 m

$\rule{190pt}{2pt}$

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

Answer 2

Step-by-step explanation:

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

Solution−

Given that,

Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide.

From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively.

Let assume that AB and CD be equal poles of height h such that AC = 80 m

Now, Let assume that P be any point on the road AC such that AP = x m and PC = (80 - x) m.

Now, In right-angle triangle APB

\begin{gathered}\rm \: tan60 \degree \: = \: \dfrac{AB}{AP} \\ \end{gathered}

tan60°=

AP

AB

\begin{gathered}\rm \: \sqrt{3} \: = \: \dfrac{h}{x} \\ \end{gathered}

3

=

x

h

\begin{gathered}\rm\implies \:h \: = \: \sqrt{3}x - - - (1) \\ \end{gathered}

⟹h=

3

x−−−(1)

Now, In right-angle triangle DCP

\begin{gathered}\rm \: tan30 \degree \: = \: \dfrac{DC}{CP} \\ \end{gathered}

tan30°=

CP

DC

\begin{gathered}\rm \: \frac{1}{ \sqrt{3} } \: = \: \dfrac{h}{80 - x} \\ \end{gathered}

3

1

=

80−x

h

On substituting the value of h from equation (1), we get

\begin{gathered}\rm \: \frac{1}{ \sqrt{3} } \: = \: \dfrac{ \sqrt{3} x}{80 - x} \\ \end{gathered}

3

1

=

80−x

3

x

\begin{gathered}\rm \: 3x = 80 - x \\ \end{gathered}

3x=80−x

\begin{gathered}\rm \: 4x = 80 \\ \end{gathered}

4x=80

\begin{gathered}\rm\implies \:x = 20 \: m \\ \end{gathered}

⟹x=20m

So, on substituting x = 20 in equation (1), we get

\begin{gathered}\rm \: h \: = \: 20 \sqrt{3} \: m \\ \end{gathered}

h=20

3

m

\begin{gathered}\rm\implies \:\boxed{ \rm{ \:Height \: of \: pole \: = \: 20 \sqrt{3} \: m \: }} \\ \end{gathered}

⟹

Heightofpole=20

3

m

Now,

Distance of the first pole AB from P = x = 20 m

Distance of the second pole CD from P = 80 - 20 = 60 m

\rule{190pt}{2pt}

\begin{gathered}\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}\end{gathered}

TrigonometryTable

∠A

sinA

cosA

tanA

cosecA

secA

cotA

0

∘

0

1

0

∞

1

∞

30

∘

2

1

2

3

3

1

2

3

2

3

45

∘

2

1

2

1

1

2

2

1

60

∘

2

3

2

1

3

3

2

2

3

1

90

∘

1

0

∞

1

∞

0