AP EAMCET · Maths · Heights and Distances
The elevation of an object on a hill is observed from a certain point in the horizontal plane through its base, to be \(30^{\circ}\). After walking 120 metres towards it on level ground the elevation is found to be \(60^{\circ}\). Then the height of the object (in metres) is :
- A 120
- B \(60 \sqrt{3}\)
- C \(120 \sqrt{3}\)
- D 60
Answer & Solution
Correct Answer
(B) \(60 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
Let \(h\) be the height of the object. In \(\triangle A C D\) \(\tan 30^{\circ}=\frac{C D}{A C}\) \(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{120+x}\) \(\Rightarrow \quad \sqrt{3} h=120+x\) ...(i) and in \(\triangle B C D\) \(\tan 60^{\circ}=\frac{C D}{B C}\)…
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