Question

A tower has obtained two stations A and B where B is East of A at a distances 100 meters. The tower is due North of A and due North-West of B. The angles of elevation of the tower from A and B are complementary. Find the height of the tower.

Answer 1

Hint: We will use the Pythagoras theorem to find the unknown sides which states that
 $ {P^2} + {B^2} = {H^2} $
Where, $ P $ is the perpendicular, $ B $ is the base and $ H $ is the hypotenuse of the triangle.
We will find the angles in the different right-angled triangles and then find their product to find the height of the tower.

Complete step-by-step answer:
We will assume the height of the tower $ PQ $ be $ h $ .
Now we will assume station $ A $ and station B to be as the point of observation.
It is given in the question that station $ B $ is in the northwest of station $ A $ .
Also point $ Q $ of the tower is in the due north of station $ A $ . The following is the schematic diagram of the tower.

It is clear from the diagram that angle $ \angle QAP = {90^ \circ } $ and $ \angle QBA = {45^ \circ } $ .
Now we will consider the triangle $ \Delta QAB $ , in which,
We have $ \angle QAB = {90^ \circ } $ and $ \angle QBA = {45^ \circ } $ . Also we know that $ QA = AB = 100\;{\rm{m}} $ .
Hence, the triangle $ \Delta QAB $ is a right-angled triangle. Therefore, we can apply Pythagoras theorem in it which states that,
 $ {P^2} + {B^2} = {H^2} $
Where, $ P $ is the perpendicular, $ B $ is the base and $ H $ is the hypotenuse of the triangle.
Now we will substitute $ 100\;{\rm{m}} $ for $ P $ and $ 100\;{\rm{m}} $ for $ B $ in the above expression to find $ H $ .
\[\begin{array}{c}
{H^2} = {\left( {100\;{\rm{m}}} \right)^2} + {\left( {100\;{\rm{m}}} \right)^2}\\
{H^2} = 2{\left( {100\;{\rm{m}}} \right)^2}\\
H = 100\sqrt 2 \;{\rm{m}}
\end{array}\]
Therefore, $ QB $ is equal to $ 100\sqrt 2 \;{\rm{m}} $ .

Now we will consider another right-angled triangle $ \Delta AQP $ .
 $ \cot \theta = \dfrac{{AQ}}{h} $
Rearranging the above expression, we get,
 $ AQ = h\cot \theta $
We will substitute $ 100\;{\rm{m}} $ for $ AQ $ in the above expression.
\[100\;{\rm{m}} = h\cot \theta \] ……(i)
Now we will consider another right-angled triangle $ \Delta BQP $ .
 $ \cot \left( {{{90}^ \circ } - \theta } \right) = \dfrac{{QB}}{h} $
Rearranging the above expression and also substituting $ \tan \theta $ for $ \cot \left( {{{90}^ \circ } - \theta } \right) $ , we get,
 $ QB = h\tan \theta $
Now we will substitute $ 100\sqrt 2 \;{\rm{m}} $ for $ QB $ in the above expression.
 $ 100\sqrt 2 \;{\rm{m}} = h\tan \theta $ ……(ii)
 Now we multiply equation (i) with (ii), we will get,
\[100\;{\rm{m}} \times 100\sqrt 2 \;{\rm{m}} = h\cot \theta \times h\tan \theta \]
Rearranging the above expression, and substituting $ \cot \theta \times \tan \theta $ for 1 we will get,
\[\begin{array}{l}
{h^2} = 100\;{\rm{m}} \times {\rm{100}}\sqrt {\rm{2}} \;{\rm{m}}\\
h = 100{\left( 2 \right)^{\dfrac{1}{4}}}\;{\rm{m}}\\
h = 118.92\;{\rm{m}}
\end{array}\]
 Hence the value of $ h $ is $ 118.92\;{\rm{m}} $ .

Note: In such types of problems, always make sure to find the values of different angles in the different right-angled triangles carefully. Also we have to use the trigonometric properties to simplify the question.