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The length of the shadow of a tower standing on level ground is found to be 2x meters longer when the sun's altitude is \[30{}^\circ \] than when it was \[45{}^\circ \]. Prove that the height of the tower is \[x\left( \sqrt{3}+~1 \right)~\] meters.

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Last updated date: 29th Mar 2024
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MVSAT 2024
Answer
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HINT: - The formula for writing tangent of an angle is

\[\tan \theta =\dfrac{Perpendicular}{\text{Base}}\] .

In this question, we will find the height of the tower in terms of x by first letting the distance between the foot of the tower to the point on the ground where the sun’s altitude is \[45{}^\circ \] . Then we will find the height of the tower in \[\Delta ABC\] and in \[\Delta ABD\] and then we will solve the two equations that are obtained to get the final solution that is the height of the tower.


Complete step-by-step solution -

As mentioned in the question, the figure would look like the below picture

seo images

By inspecting the figure, we have can write that

Let the height of tower AB which is taken as h.

Also, let the distance between the foot of the tower and the point at which the sun has an altitude \[{{45}^{\circ }}\] .

Now, the angles of elevations that are the altitudes of sun are \[{{30}^{\circ }}\] and \[{{45}^{\circ }}\] at two different points on the ground.

Also, we know that

\[\angle ACB={{30}^{\circ }}\ and\ \angle ADB={{45}^{\circ }}\]

(As given in the question)

Now, using the formula for tangent in the \[\Delta ABC\] , we get

\[ \tan \theta =\dfrac{Perpendicular}{\text{Base}} \]

 \[ \tan {{30}^{\circ }}=\dfrac{h}{2x+y} \]

 \[ \dfrac{1}{\sqrt{3}}=\dfrac{h}{2x+y} \]

 \[ h=\dfrac{2x+y}{\sqrt{3}} \]

Similarly, using the formula for tangent in the \[\Delta ABD\] , we get

\[ \tan \theta =\dfrac{Perpendicular}{\text{Base}} \]

 \[ \tan {{45}^{\circ }}=\dfrac{h}{y} \]

 \[ 1=\dfrac{h}{y} \]

 \[ y=h \]

Now, using the value of y, we can find the value of h in terms of x and hence, we get

\[ \dfrac{2x+y}{\sqrt{3}}=h \]

\[ \dfrac{2x+h}{\sqrt{3}}=h \]

 \[ \dfrac{2x}{\sqrt{3}}+\dfrac{h}{\sqrt{3}}=h \]

 \[ \dfrac{2x}{\sqrt{3}}=h-\dfrac{h}{\sqrt{3}} \]

 \[ h\left( \sqrt{3}-1 \right)=2x \]

 \[ h=\dfrac{2x}{\left( \sqrt{3}-1 \right)} \]

 \[ h=\dfrac{2x\left( \sqrt{3}+1 \right)}{\left( \sqrt{3}-1 \right)\cdot \left( \sqrt{3}+1 \right)} \]

 \[ h=\dfrac{2x\left( \sqrt{3}+1 \right)}{{{\left( \sqrt{3} \right)}^{2}}-1}=\dfrac{2x\left( \sqrt{3}+1 \right)}{3-1}=\dfrac{2x\left( \sqrt{3}+1 \right)}{2}=x\left( \sqrt{3}+1 \right) \]

 \[ h=x\left( \sqrt{3}+1 \right) \]

Hence, the height of the tower is h which is equal to \[x\left( \sqrt{3}+1 \right)\] m.


NOTE: - The figure in this question is very tricky and is difficult to visualize it at first. Hence, the students can make an error while drawing the figure and then end up doing a mistake and they would get to the correct solution.


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