From the top of building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be ${{30}^{0}}$ and ${{60}^{0}}$ respectively. Find
(i) the horizontal distance between AB and CD.
(ii) the height of the lamp post.
(iii) the difference between the heights of the building and the lamp post.
VerifiedHint: For solving this problem first we will draw the geometrical figure as per the given data. After that, we will use the basic formula of trigonometry $\tan \theta =\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)}$ . Then, we will solve correctly to get the correct answer for each part.
Complete step-by-step answer:
Given:
It is given that, from the top of building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be ${{30}^{0}}$ and ${{60}^{0}}$ respectively. And we have to find the answer of the following parts:
(i) the horizontal distance between AB and CD.
(ii) the height of the lamp post.
(iii) the difference between the heights of the building and the lamp post.
Now, first, we will draw a geometrical figure as per the given data. For more clarity look at the figure given below:
In the above figure, AB represents the 60 m length of the building and CD represents the length of the vertical lamp post. Moreover, as per the given data angle depression of the top and bottom of the lamp post CD from the point A is equal to $\angle CAE={{30}^{0}}$ and $\angle DAE={{60}^{0}}$ respectively.
Now, as points E, C and D lie on a vertical line and DC is vertical and DE will be equal to the length of the segment AB. Then,
\[ CD+CE=DE \]
\[ \Rightarrow CD=AB-CE.......................\left( 1 \right) \\]
Now, we consider $\Delta DAE$ in which $\angle DEA={{90}^{0}}$ , $DE=60$ is the length of the perpendicular, AE is the length of the base and $\angle DAE={{60}^{0}}$ . Then,
\[ \tan \left( \angle DAE \right)=\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)} \]
\[ \Rightarrow \tan {{60}^{0}}=\dfrac{DE}{AE} \]
\[ \Rightarrow \sqrt{3}=\dfrac{60}{AE} \]
\[ \Rightarrow AE=\dfrac{60}{\sqrt{3}} \]
\[ \Rightarrow AE=20\sqrt{3}.........................\left( 2 \right) \]
Now, as point E, C and D lie on a vertical line segment DC which is parallel to segment AB so, AE will be equal to the length of the segment DB. Then,
Horizontal distance between AB and CD $=AE=DB=20\sqrt{3}$ metres.
Now, we consider $\Delta CAE$ in which $\angle CEA={{90}^{0}}$ , CE is the length of the perpendicular, $AE=20\sqrt{3}$ is the length of the base and $\angle CAE={{30}^{0}}$ . Then,
\[ \tan \left( \angle CAE \right)=\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)} \]
\[ \Rightarrow \tan {{30}^{0}}=\dfrac{CE}{AE} \]
\[ \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{CE}{20\sqrt{3}} \]
\[ \Rightarrow CE=20.........................\left( 2 \right) \]
Now, from the above result, it is evident that the difference between the heights of the building and the lamp post will be $CE=20$ metres.
Now, put $CE=20$ and $AB=60$ in equation (1). Then,
\[ CD=AB-CE \]
\[ \Rightarrow CD=60-20 \]
\[ \Rightarrow CD=40 \]
Now, from the above result, it is evident that the height of the lamp post CD is equal to 40 metres.
Now, from the above results we conclude the following results:
(i) Horizontal distance between AB and CD will be equal to $20\sqrt{3}\approx 34.641$ metres.
(ii) Height of the lamp post will be equal to 40 metres.
(iii) Difference between the heights of the building and the lamp post will be equal to 20 metres.
Note: Here, the student should first try to understand what is asked in the problem. After that, we should try to draw the geometrical figure as per the given data and proceed stepwise and, while making the figure we should remember that angle of depression of the top and bottom of the lamp post from the top of the building is given. Moreover, we should apply the basic formula of trigonometry properly without any error and avoid calculation mistakes while solving to get the correct answer.
