Question

A lamp post stands at the centre of the circular park. Let P and Q be the two points on the boundary such that PQ subtends an angle ${{90}^{\circ }}$at the foot of the lamp post and the angle of the elevation at the top of the lamp post from point P is ${{30}^{\circ }}$. If PQ=30m, then find the height of the lamp post.
(a)$5\sqrt{6}m$
(b)$4\sqrt{6}m$
(c)$6\sqrt{5}m$
(d)$8\sqrt{3}m$

Answer 1

Hint: Firstly, we will use the formula of the Pythagoras theorem which states that in the right angled triangle that square of the hypotenuse is equal to the sum of the square of the base and height of that triangle. Then, we will use the concept of trigonometry in which the formula is there as $\tan \theta =\dfrac{P}{B}$. Then, by substituting the value of the angle and base of the triangle, we get the height of the lamp post.

Complete step-by-step answer:
In this question, first we will draw the figure of the circular park as circle with two radiuses as OP and OQ which is making an angle of ${{90}^{\circ }}$with each other and OR represents the height of the lamp post which is centred at the circular park.

Here, the radius of the circle OP and OQ are equal as:
OP=OQ
Also, we are given with the side as PQ=30m.
Since, we know by Pythagora's theorem the formula of the right angled triangle that square of the hypotenuse is equal to the sum of the square of the base and height of that triangle.
Applying Pythagoras theorem in the triangle OPQ, substitute PQ=30 and use OP=OQ, we get
$\begin{align}
  & P{{Q}^{2}}=O{{P}^{2}}+O{{Q}^{2}} \\
 & \Rightarrow {{30}^{2}}=O{{P}^{2}}+O{{P}^{2}} \\
 & \Rightarrow 900=2O{{P}^{2}} \\
 & \Rightarrow 450=O{{P}^{2}} \\
 & \Rightarrow OP=\sqrt{450} \\
 & \Rightarrow OP=15\sqrt{2} \\
\end{align}$
Now, to get the height of the lamp post which is OR is calculated by using the formula of trigonometry as:
$\tan \theta =\dfrac{P}{B}$
Where P is perpendicular height and B is base of that post for this question.
Now substituting the values of the angle which is ${{30}^{\circ }}$ from the figure and base OP value which is calculated above to get the height(H) of the lamp post as:
$\begin{align}
  & \tan P=\dfrac{H}{OP} \\
 & \Rightarrow \tan {{30}^{\circ }}=\dfrac{H}{15\sqrt{2}} \\
 & \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{H}{15\sqrt{2}} \\
 & \Rightarrow H=\dfrac{15\sqrt{6}}{3} \\
 & \Rightarrow H=5\sqrt{6} \\
\end{align}$
So, the height of the lamp post is $5\sqrt{6}m$.
Hence, option (a) is correct.

Note: The most common mistake in this type of the questions is on considering the base and perpendicular of the triangle for using the formula of trigonometry as $\tan \theta =\dfrac{P}{B}$. We should be very careful in considering the base and perpendicular as the base is the side at the angle is given which is ${{30}^{\circ }}$. So, the angle is always considered between the base and hypotenuse and left outside is always perpendicular. If we interchange the value of the base and perpendicular, we get the answer as:
$\begin{align}
  & \tan P=\dfrac{OP}{H} \\
 & \Rightarrow \tan {{30}^{\circ }}=\dfrac{15\sqrt{2}}{H} \\
 & \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{15\sqrt{2}}{H} \\
 & \Rightarrow H=15\sqrt{6} \\
\end{align}$
So, we get the wrong answer for the height of the lamp post as $15\sqrt{6}m$.