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Question: Two poles are \('a'\) metres apart and the height of one is double of the other. If from the middle ...

Two poles are a'a' metres apart and the height of one is double of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the smaller is.
(a) 2a\sqrt{2}a metres
(b) a22\dfrac{a}{2\sqrt{2}} metres
(c) a2\dfrac{a}{\sqrt{2}} metres
(d) 2a2a metres

Explanation

Solution

Hint: 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 tanθ=(length of the perpendicular)(length of the base)\tan \theta =\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)} and tan(π2θ)=cotθ=1tanθ\tan \left( \dfrac{\pi }{2}-\theta \right)=\cot \theta =\dfrac{1}{\tan \theta } . Then, we will solve correctly to get the height of the smaller pole.

Complete step-by-step answer:

Given:

It is given that there are two poles that are a'a' meters apart and the height of one is double of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary. And we have to find the height of the smaller pole.

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 BA represents the height of the larger pole, DC represents the height of the smaller pole and AC represents the distance between their feet, so AC=aAC=a metres. And E is the midpoint of the AC so, AE=EC=a2AE=EC=\dfrac{a}{2} metres. Moreover, α\alpha and β\beta are the angle of elevations of the topmost point of the larger and smaller poles respectively and it is given that these angles are complementary so, the value of α+β=π2\alpha +\beta =\dfrac{\pi }{2} .

Now, consider ΔABE\Delta ABE in which BAE=900\angle BAE={{90}^{0}} , AE is equal to the length of the base, BA is equal to the length of the perpendicular and BEA=α\angle BEA=\alpha . Then,

tan ( \angle BEA)=(length of the perpendicular)(length of the base)\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)}

tanα=BAAE\Rightarrow \tan \alpha =\dfrac{BA}{AE}

tanα=2(BA)a............................(1)\Rightarrow \tan \alpha =\dfrac{2\left( BA \right)}{a}............................\left( 1 \right)

Now, consider ΔCDE\Delta CDE in which DCE=900\angle DCE={{90}^{0}} , EC is equal to the length of the base, DC is equal to the length of the perpendicular and DEC=β\angle DEC=\beta . Then,

tan ( \angle DEC )=(length of the perpendicular)(length of the base)\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)}

tanβ=DCEC\Rightarrow \tan \beta =\dfrac{DC}{EC}

tanβ=2(DC)a............................(2)\Rightarrow \tan \beta =\dfrac{2\left( DC \right)}{a}............................\left( 2 \right)

Now, as it is given that from the middle point of the line joining their feet, an observer finds the angular elevation of their tops to be complementary. Then,

α+β\alpha + \beta = π2\dfrac{\pi }{2}

α=π2β\Rightarrow \alpha =\dfrac{\pi }{2}-\beta

tanα=tan(π2β)\Rightarrow \tan \alpha =\tan \left( \dfrac{\pi }{2}-\beta \right)

tanα=cotβ\Rightarrow \tan \alpha =\cot \beta

tanα=1tanβ\Rightarrow \tan \alpha =\dfrac{1}{\tan \beta }

tanαtanβ=1\Rightarrow \tan \alpha \tan \beta =1

Now, substitute tanα=2(BA)a\tan \alpha =\dfrac{2\left( BA \right)}{a} from equation (1) and tanβ=2(DC)a\tan \beta =\dfrac{2\left( DC \right)}{a} from equation (2) in the above equation. Then,

tan α\alpha tan β\beta =1

2(BA)a×2(DC)a=1\Rightarrow \dfrac{2\left( BA \right)}{a}\times \dfrac{2\left( DC \right)}{a}=1

(BA)×(DC)=a24\Rightarrow \left( BA \right)\times \left( DC \right)=\dfrac{{{a}^{2}}}{4}

Now, as BA represents the height of the taller person and DC represents the height of the shorter person. And it is given that the height of one is double that of the other so, we can substitute BA=2(DC)BA=2\left( DC \right) in the above equation. Then,

( BA )×\times ( DC)=a24\dfrac{{{a}^{2}}}{4}

2(DC)2=a24\Rightarrow 2{{\left( DC \right)}^{2}}=\dfrac{{{a}^{2}}}{4}

(DC)2=a28\Rightarrow {{\left( DC \right)}^{2}}=\dfrac{{{a}^{2}}}{8}

DC=a22\Rightarrow DC=\dfrac{a}{2\sqrt{2}}

Now, from the result, we conclude that the length of DC will be equal to a22\dfrac{a}{2\sqrt{2}} metres.

Thus, the height of the smaller tower will be a22\dfrac{a}{2\sqrt{2}} metres.

Hence, (b) is the correct option.

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. Moreover, we should apply the basic formula of trigonometry properly without any error and avoid calculation mistakes while solving to get the correct answer and then select the correct option.