question_answer

Two stations due south of a leaning tower, which leans towards the north, are at distances a and b from its foot. If x and y are the angles of elevations of the top of the tower from these stations then its  inclination \[\theta \] to the horizontal is given by \[\cot \theta =\]

A)  \[\frac{b\,\cot x+a\,\cot y}{b-a}\]

B)  \[\frac{b\,\cot x-a\,\cot y}{b-a}\]

C)  \[\frac{a\,\cot x-b\,\cot y}{b+a}\]

D)  None of these

Correct Answer: B

Solution :

Let CD, A and B represent the leaning tower and the two given points. \[AC=a,\text{ }BC=b\] \[\angle DCM=\theta ,\,\angle DAM=x,\,\angle DBM=y\] Let DM = h and CM = z. In \[\Delta \,DCM,\,\,\frac{DM}{CM}=\tan \theta \] \[\Rightarrow \] \[\frac{h}{z}=\tan \theta \Rightarrow z=h\cot \theta \]              ??(1) In  \[\Delta \,DAM,\frac{DM}{AM}=\tan x\] \[\Rightarrow \]  \[\frac{h}{a+z}=\tan x\Rightarrow a+z=h\,\cot x.\] \[\Rightarrow \]\[a=h\cot x-z=h\cot x-h\cot \theta \] (using (1)) \[\Rightarrow \] \[a=h(\cot \,x-\cot \theta )\]     ?..(2) In \[\Delta DBM,\,\,\frac{DM}{BM}=\tan y\Rightarrow b+z=h\cot y\] \[\Rightarrow \] \[b=h\cot y-z=h\cot y-h\cot \theta \] \[\Rightarrow \] \[b=h(\cot y-\cot \theta )\] ?(3)[From eq. (1)] Dividing eq. (2) by (3), we get \[\frac{a}{b}=\frac{h\,(\cot x-\cot \theta )}{h(\cot y-\cot \theta )}\] \[\Rightarrow \] \[a\cot y-a\cot \theta =b\cot x-b\cot \theta \] \[\Rightarrow \] \[(b-a)\cot \theta =b\cot x-a\cot y\] \[\Rightarrow \] \[\cot \theta =\frac{b\cot x-a\cot y}{b-a}\]