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Question: The length of the shadow of a tower standing on level ground is found to be \(2x\) metres longer whe...

The length of the shadow of a tower standing on level ground is found to be 2x2x metres longer when the sun’s elevation is 300{{30}^{0}} than when it was 450{{45}^{0}} . The height of the tower in metres is.
(a) (3+1)x\left( \sqrt{3}+1 \right)x
(b) (3−1)x\left( \sqrt{3}-1 \right)x
(c) 23x2\sqrt{3}x
(d) 32x3\sqrt{2}x

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 tan⁡θ=(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)} . Then, we will solve correctly to get the height of the tower and select the correct option.

Complete step-by-step answer:
Given:
It is given that the length of the shadow of a tower standing on level ground is found to be 2x2x metres longer when the sun’s elevation is 300{{30}^{0}} than when it was 450{{45}^{0}} . And we have to find the height of the tower.
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 tower, DA represents the length of the shadow of the tower when the elevation of the sun is equal to ∠BDA=β=450\angle BDA=\beta ={{45}^{0}} and CA represents the length of the shadow of the tower when the elevation of the sun is equal to ∠BCA=α=300\angle BCA=\alpha ={{30}^{0}} .
Now, as it is given that length of the shadow of a tower standing on level ground is found to be 2x2x metres longer when the sun’s elevation is 300{{30}^{0}} than when it was 450{{45}^{0}} . Then,
CA=DA+2x...................(1)CA=DA+2x...................\left( 1 \right)
Now, consider ΔBDA\Delta BDA in which ∠BAD=900\angle BAD={{90}^{0}} , DA is equal to the length of the base, BA is equal to the length of the perpendicular and ∠BDA=β=450\angle BDA=\beta ={{45}^{0}}. Then,
tan⁡(∠BDA)=(length of the perpendicular)(length of the base) ⇒tan⁡450=BADA ⇒1=BADA ⇒DA=BA................................(2) \begin{aligned} & \tan \left( \angle BDA \right)=\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)} \\\ & \Rightarrow \tan {{45}^{0}}=\dfrac{BA}{DA} \\\ & \Rightarrow 1=\dfrac{BA}{DA} \\\ & \Rightarrow DA=BA................................\left( 2 \right) \\\ \end{aligned}
Now, consider ΔBCA\Delta BCA in which ∠BAC=900\angle BAC={{90}^{0}} , CA is equal to the length of the base, BA is equal to the length of the perpendicular and ∠BCA=α=300\angle BCA=\alpha ={{30}^{0}}. Then,
tan⁡(∠BCA)=(length of the perpendicular)(length of the base) ⇒tan⁡300=BACA ⇒13=BACA ⇒CA=3(BA)................................(3) \begin{aligned} & \tan \left( \angle BCA \right)=\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)} \\\ & \Rightarrow \tan {{30}^{0}}=\dfrac{BA}{CA} \\\ & \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{BA}{CA} \\\ & \Rightarrow CA=\sqrt{3}\left( BA \right)................................\left( 3 \right) \\\ \end{aligned}
Now, substitute DA=BADA=BA from equation (2) and CA=3(BA)CA=\sqrt{3}\left( BA \right) from equation (3) into equation (1). Then,
CA=DA+2x ⇒3(BA)=BA+2x ⇒BA(3−1)=2x ⇒BA=2x(3−1) ⇒BA=2x(3−1)×(3+1)(3+1) \begin{aligned} & CA=DA+2x \\\ & \Rightarrow \sqrt{3}\left( BA \right)=BA+2x \\\ & \Rightarrow BA\left( \sqrt{3}-1 \right)=2x \\\ & \Rightarrow BA=\dfrac{2x}{\left( \sqrt{3}-1 \right)} \\\ & \Rightarrow BA=\dfrac{2x}{\left( \sqrt{3}-1 \right)}\times \dfrac{\left( \sqrt{3}+1 \right)}{\left( \sqrt{3}+1 \right)} \\\ \end{aligned}
Now, as we know that (a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} so, we can write (3−1)(3+1)=(3)2−12\left( \sqrt{3}-1 \right)\left( \sqrt{3}+1 \right)={{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}} in the above equation. Then,
BA=2x(3−1)×(3+1)(3+1) ⇒BA=2x(3+1)((3)2−12) ⇒BA=2x(3+1)(3−1) ⇒BA=2x(3+1)2 ⇒BA=x(3+1) \begin{aligned} & BA=\dfrac{2x}{\left( \sqrt{3}-1 \right)}\times \dfrac{\left( \sqrt{3}+1 \right)}{\left( \sqrt{3}+1 \right)} \\\ & \Rightarrow BA=\dfrac{2x\left( \sqrt{3}+1 \right)}{\left( {{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}} \right)} \\\ & \Rightarrow BA=\dfrac{2x\left( \sqrt{3}+1 \right)}{\left( 3-1 \right)} \\\ & \Rightarrow BA=\dfrac{2x\left( \sqrt{3}+1 \right)}{2} \\\ & \Rightarrow BA=x\left( \sqrt{3}+1 \right) \\\ \end{aligned}
Now, from the above result, we conclude that the length of the BA will be x(3+1)x\left( \sqrt{3}+1 \right) metres.
Thus, the height of the tower will be x(3+1)x\left( \sqrt{3}+1 \right) metres.
Hence, (a) 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.