question_answer

Two villages are 2 kms apart. If the angles of depression of these villages when observed from a plane are found to be \[45{}^\circ \] and \[60{}^\circ \] respectively, then height of the plane is

A) \[(3+\sqrt{3})\,km\]

B) \[(3-\sqrt{3})\,km\]

C) \[2\sqrt{3}\,km\]

D) \[3\sqrt{3}\,km\]

Correct Answer: B

Solution :

 Let \[PQ=h\,\,m\] be a plane From right angled \[\Delta \,PAQ,\]                                 \[\tan {{45}^{o}}=\frac{PQ}{AQ}\]                 \[\therefore \]  \[1=\frac{h}{x}\]                 or            \[x=h\]                                 ??(i) Again from right angled \[\Delta PBQ,\]                 ©                 \[\tan {{60}^{o}}=\frac{PQ}{QB}\] or            \[\sqrt{3}=\frac{h}{2-x}\] From equation (i), \[\sqrt{3}=\frac{h}{2-h}\] or            \[2\sqrt{3}-\sqrt{3}\,h=h\] or            \[h+\sqrt{3}h=2\sqrt{3}\] or            \[h(1+\sqrt{3})=2\sqrt{3}\] or          \[h=\frac{2\sqrt{3}}{1+\sqrt{3}}=\frac{2\sqrt{3}}{1+\sqrt{3}}\times \frac{1-\sqrt{3}}{1-\sqrt{3}}\]                 \[=\frac{2\sqrt{3}-6}{1-3}=\frac{-2(3-\sqrt{3})}{-2}\]                 \[=(3-\sqrt{3})km\]