Question

The angle of elevation of the top of a vertical tower from two points at distances $ a $ and $ b $ $ (a > b) $ from the base and in the same line with it, are complimentary. If $ \theta $ is the angle subtended at the top of the tower by the line joining these points, then $ \sin \theta $ is equal to
A. $ \dfrac{{a + b}}{{a - b}} $
B. $ \dfrac{{a - b}}{{a + b}} $
C. $ \dfrac{{(a - b)b}}{{a + b}} $
D. $ \dfrac{{a - b}}{{(a + b)b}} $

Answer 1

Hint: As we can see that in the above question we have to find the value of $ \sin \theta $ which means that there is use of trigonometric application and identities to get the required answer. We will first draw the diagram according to the data given in the question . We will assume one of the angles as $ \phi $ as it cannot be $ \theta $ because it is the angle subtended at the top of the tower by the line joining the points.

Complete step-by-step answer:
Let us first draw the diagram according to the question:

In the above figure we have $ \angle BAD = \theta $ ( $ \theta $ is the angle subtended at the top of the tower by the line joining these points).
Let us assume $ \angle CBA = \phi ,\angle ADC = 90 - \phi $ . Also we have $ CD = a $ and $ BC = b $ .
Now in triangle CAD , we have $ \tan \phi $ , so we have $ \tan \phi = \dfrac{p}{b} $ , here the perpendicular is AC and the base is $ BC = b $ . So by putting the value we can write $ \tan \phi = \dfrac{{AC}}{b} $ .
Again in triangle CAD we can write $ \tan (90 - \phi ) = \dfrac{{AC}}{a} $ .
Now we know that $ \tan (90 - \theta ) = \cot \theta $ , so it can be written as $ \cot \phi = \dfrac{{AC}}{a} $ .
 We know the trigonometric identity that $ \cot \theta = \dfrac{1}{{\tan \theta }} $ , so by this we can write from the above expressions , $ \dfrac{{AC}}{a} = \dfrac{1}{{\dfrac{{AC}}{b}}} \Rightarrow \dfrac{{AC}}{a} = \dfrac{a}{{AC}} $ .
By cross multiplication it can be written as $ A{C^2} = ab $ or $ AC = \sqrt {ab} $ .
Now we can write $ \sin \phi = \dfrac{p}{h} $ , here we have perpendicular is $ AC = \sqrt {ab} $ and hypotenuse $ AB $ can be written as $ A{C^2} + B{C^2} $ , we have $ AC = \sqrt {ab} $ and $ BC = b $ .
So we have hypotenuse \[{h^2} = {(\sqrt {ab} )^2} + {b^2}\], by removing the square of hypotenuse we have i.e. $ h = \sqrt {ab + {b^2}} $ .
After putting these values back we have $ \sin \phi = \sqrt {\dfrac{{ab}}{{ab + {b^2}}}} $ .
By squaring both the sides we have $ {\sin ^2}\phi = {\left( {\sqrt {\dfrac{{ab}}{{ab + {b^2}}}} } \right)^2} $ . Since squaring the square root removes it, we have $ \dfrac{{ab}}{{ab + {b^2}}} $ . We can take the common factor out in the denominator, so we have $ \dfrac{{ab}}{{b(a + b)}} = \dfrac{a}{{a + b}} $ .
From the above figure and by applying trigonometric identity we can write $ \sin \theta = - 1 + 2{\sin ^2}\phi $ .
By putting the value in the identity we have $ - 1 + \dfrac{{2a}}{{a + b}} $ , On solving we have $ \dfrac{{ - a - b + 2a}}{{a + b}} \Rightarrow \dfrac{{a - b}}{{a + b}} $ .
Hence the correct option is (b) $ \dfrac{{a - b}}{{a + b}} $ .
So, the correct answer is “Option B”.

Note: We should note that in the above solution we have used the trigonometric identity that $ \sin \theta = - \sin \left( {90 - \phi } \right) = - \cos 2\phi $ . Then there is another identity which says that $ \cos 2\theta $ can be written as $ 1 - 2{\sin ^2}\theta $ , so we can write here $ - \cos 2\phi = - 1 + 2{\sin ^2}\phi $ . In triangle CAD we have perpendicular $ = AC $ and base $ = a $ .