Question

Two pillars of equal heights stand on either side of a road which is 150 m wide. At a point on the road between the pillars, the angle of elevation of the tops of the pillars are 600 and 300 . Find the height of each pillar.

Answer 1

Appropriate Question :-

Two pillars of equal heights stand on either side of a road which is 150 m wide. At a point on the road between the pillars, the angle of elevation of the tops of the pillars are 60° and 30° . Find the height of each pillar.

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

Let assume that AB and CD are two pillars of height 'h' on the either side of a road which is 150 m wide.

Let P be the point on the road at a distance of 'x' m from point B such that the angle of elevation of the top of the pillars are 60° and 30°

Now, In right triangle ABP

$\rm \: tan60\degree \: = \: \frac{AB}{BP}$

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

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

Now, In right triangle CDP

$\rm \: tan30\degree \: = \: \frac{CD}{DP}$

$\rm \: \dfrac{1}{ \sqrt{3} } = \dfrac{h}{150 - x}$

$\rm \: \dfrac{1}{ \sqrt{3} } = \dfrac{ \sqrt{3} x}{150 - x}$

$\rm \: 3x = 150 - x$

$\rm \: 3x + x = 150$

$\rm \: 4x = 150$

$\rm\implies \:x = 37.5 \: m$

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

$\rm \: h = 37.5 \sqrt{3}$

$\rm \: h = 37.5 \times 1.732$

$\rm\implies \:h \: = \: 64.95 \: m$

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

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

Answer 2

Step-by-step explanation:

Width of the road =150 m

Angle of the first pillars =60°

Angle of the first pillars =30°

Let AB and CD be two pillars.

Height of pillars =h meter

Observation point on the road is P.

In triangle PAB,

$\\ \rm \tan 60^{\circ}=\frac{ AB }{ AP }$

$\[ \begin{array}{l} \displaystyle \rm \Rightarrow \sqrt{3}=\frac{h}{x} \\ \\ \displaystyle \rm\Rightarrow \sqrt{3} x=h \end{array} \]$

In triangle PCD

$\[ \begin{array}{l} \displaystyle \rm\tan 30^{\circ}=\frac{ CD }{ CP } \\ \\ \displaystyle \rm \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{150-x} \\ \\ \displaystyle \rm\Rightarrow h \sqrt{3}=150-x \ldots .(2) \end{array} \]$

Put the value of h from equation (1) in equation (2),

$\pmb{\[ \begin{array}{l} \displaystyle \rm \Rightarrow x \sqrt{3} \times \sqrt{3}=150-x \\ \\ \displaystyle \rm\Rightarrow 3 x=150-x \\ \\\displaystyle \rm \Rightarrow x=37.5 \end{array} \]}$

Put this value in equation (1),

$\pmb{\[ \begin{array}{l}\displaystyle \rm \Rightarrow \sqrt{3} \times 37.5=h \\ \\ \displaystyle \rm\Rightarrow h=64.95 \end{array} \]}$

Thus the required point is at the distance of 37.5 m from the first pillar. The height of the pillars is 64.95 m .