Question

At a distance \[a\] from the foot of a tower \[AB\] of known height \[b\] a flagstaff \[BC\] and the tower subtend equal angles, height of the flagstaff is
A) \[\dfrac{{{a^2} + {b^2}}}{{{a^2} - {b^2}}}\]
B) \[\dfrac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}\]
C) \[\dfrac{{a\left( {{a^2} - {b^2}} \right)}}{{{a^2} + {b^2}}}\]
D) \[\dfrac{{b\left( {{a^2} + {b^2}} \right)}}{{{a^2} - {b^2}}}\]

Answer 1

Hint:
Here we will draw the figure of the tower and flagstaff according to the data given in the question. Then we will use the trigonometric function \[\tan \theta \] to get the relation between the height of the tower and the base. Again, we will find \[\tan 2\theta \] to get the relation between the total height of the tower and flagstaff and the base. We will further substitute the value of \[\tan \theta \] in trigonometric property of \[\tan 2\theta \] to get the value of the height of the flagstaff.

Complete step by step solution:
Firstly, we will draw the figure according to the data given in the question.

We know that, \[\angle APB = \angle BPA = \theta \] also \[AB = b\].
Now we will use the trigonometric function \[\tan \theta \] to get the relation between the height of the tower and the base. Therefore, in \[\Delta ABP\], \[\angle BAP = 90^\circ \].
Now using the formula for \[\tan \theta \] in \[\Delta ABP\], we get
\[\tan \theta = \dfrac{{AB}}{{AP}} = \dfrac{b}{a}\]…………………. \[\left( 1 \right)\]
Now we will find the trigonometric function \[\tan 2\theta \] to get the relation between the total height of the tower and flagstaff and the base. Therefore, in \[\Delta CAP\], \[\angle CAP = 90^\circ \].
Now using the formula for \[\tan 2\theta \] in \[\Delta CAP\], we get
\[\tan 2\theta = \dfrac{{AC}}{{AP}} = \dfrac{{b + BC}}{a}\]……………… \[\left( 2 \right)\]
Now, we know from the property of the trigonometry that \[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\].
So, substituting \[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\] in the above equation, we get
\[ \Rightarrow \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = \dfrac{{b + BC}}{a}\]
Now we will put the value of the \[\tan \theta \]from the equation \[\left( 1 \right)\], we get
\[ \Rightarrow \dfrac{{2\left( {\dfrac{b}{a}} \right)}}{{1 - {{\left( {\dfrac{b}{a}} \right)}^2}}} = \dfrac{{b + BC}}{a}\]
Now we will simplify the above equation to get the value of \[BC\] which is the height of the flagstaff. Therefore, we get
\[ \Rightarrow \dfrac{{2ab}}{{{a^2} - {b^2}}} = \dfrac{{b + BC}}{a}\]
Multiplying \[a\] on both sides, we get
\[ \Rightarrow b + BC = \dfrac{{2{a^2}b}}{{{a^2} - {b^2}}}\]
Subtracting \[b\] on both sides, we get
\[ \Rightarrow BC = \dfrac{{2{a^2}b}}{{{a^2} - {b^2}}} - b\]
Taking LCM, we get
\[ \Rightarrow BC = \dfrac{{2{a^2}b - b\left( {{a^2} - {b^2}} \right)}}{{{a^2} - {b^2}}}\]
Using distributive property, we get
\[ \Rightarrow BC = \dfrac{{2{a^2}b - {a^2}b + {b^3}}}{{{a^2} - {b^2}}} = \dfrac{{{a^2}b + {b^3}}}{{{a^2} - {b^2}}} = \dfrac{{b\left( {{a^2} + {b^2}} \right)}}{{{a^2} - {b^2}}}\]
Simplifying the terms, we get
\[\begin{array}{l} \Rightarrow BC = \dfrac{{{a^2}b + {b^3}}}{{{a^2} - {b^2}}}\\ \Rightarrow BC = \dfrac{{b\left( {{a^2} + {b^2}} \right)}}{{{a^2} - {b^2}}}\end{array}\]
Hence, the height of the flagstaff is \[\dfrac{{b\left( {{a^2} + {b^2}} \right)}}{{{a^2} - {b^2}}}\].

So, option D is the correct option.

Note:
Here we have to note that only by trigonometric functions we can get the relationship between the height and the distance of the base. There are different properties like
\[\begin{array}{l}\sin 2\theta = 2\sin \theta \cos \theta \\\cos 2\theta = 2{\cos ^2}\theta - 1 = 1 - 2{\sin ^2}\theta = {\cos ^2}\theta - {\sin ^2}\theta \\\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\end{array}\]
We should know that the trigonometric functions are only applicable for the right angled triangle.