question_answer

A tower is situated on horizontal plane. From two points, the line joining three points passes through the base and which are \[a\]and \[b\]distance from the base. The angle of elevation of the top are \[\alpha \]and \[90{}^\circ -\alpha \]and \[\theta \]is that angle which two points joining the line makes at the top, the height of tower will be [UPSEAT 1999]

A) \[\frac{a+b}{a-b}\]

B) \[\frac{a-b}{a+b}\]

C) \[\sqrt{ab}\]

D) \[{{(ab)}^{1/3}}\]

Correct Answer: C

Solution :

Let there are two points C and D on horizontal line passing from point B of the base of the tower AB.  The distance of these points are b and a from B respectively. \[\therefore BD=a\,\,\text{and}\,\,BC=b\]. \[\because \] line CD, on the top of tower A subtends an angle \[\theta \], \[\therefore \,\,\angle CAD=\theta \] According to question, on point C  and D, the elevation of top are \[\alpha \] and \[{{90}^{o}}-\alpha \]. \[\therefore \,\,\,\,\angle BCA\,\,\,\,=\,\,\,\alpha \] and \[\,\,\angle BDA=\]\[{{90}^{o}}-\alpha \]. In \[\Delta ABC,\] \[AB=BC\tan \alpha \]  \[=b\tan \alpha \]          ?.. (i) and in\[\Delta ABD\], AB = BD\[\tan ({{90}^{o}}-\alpha )\]\[=a\cot \alpha \] ?..(ii) Multiplying equation (i) and (ii), \[{{(AB)}^{2}}=(b\tan \alpha )(a\cot \alpha )=ab\], \[\therefore \,AB=\sqrt{(ab)}\].