Question

a contractor construct a vertical pillar at a horizontal distance of 300m from a fixed point it was decided that the angel of elevation of the top of complete pillar from that point to be 60 degree. He finished the job by making a pillar such that the angle of top of it is 30 degree. Find the height of the pillar to be increased as per the terms of contract.

Answer 1

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Answer 2

Answer:

The height of the pillar to be increased as per the terms of contract is 346.4 m

Step-by-step explanation:

Given the distance between the pillar and the fixed point (D) is AD = 300 m

Let consider the height of the pillar to be constructed $AB = h_{1}$

In $\Delta ABD$ is a right angle triangle and $\angle ADB=60^{\circ}$

therefore,

$\frac{AB}{AD}=tan60^{\circ}\\\\\Rightarrow{AB}=\sqrt{3}\times{AD}\\\\\Rightarrow{h_{1}}=\sqrt{3}\times300$

Now consider the actual height of the pillar = AC=$h_{2}$

And x is the height of the pillar to be increased to meet the contract terms.

therefore

$h_{2}=h_{1}-x$

In $\Delta ACD$ is a right angle triangle and $\angle ADC=30^{\circ}$

$\frac{AC}{AD}=tan30^{\circ}\\\\\Rightarrow{AC}=\frac{1}{\sqrt{3}} \times{AD}\\\\\Rightarrow{h_{2}}=\frac{1}{\sqrt{3}}\times300\\\\\Rightarrow{h_{1}-x}=\frac{1}{\sqrt{3}}\times300\\\\\Rightarrow{x}=h_{1}-\frac{1}{\sqrt{3}}\times300\\\\\Rightarrow{x}=\sqrt{3}\times300-\frac{1}{\sqrt{3}}\times300\\\\\Rightarrow{x}=300(\sqrt{3}-\frac{1}{\sqrt{3}})\\\\\Rightarrow{x}=300(\frac{3-1}{\sqrt{3}})\\\\\Rightarrow{x}=300(\frac{2}{\sqrt{3}})\\\\\Rightarrow{x}=\frac{600}{\sqrt{3}}\\\\\Rightarrow{x}=346.4\text{ m}$

The height of the pillar to be increased as per the terms of contract is 346.4 m