Question

The angle of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m.

Answer 1

Hint: To prove this we must have knowledge of trigonometric ratios. At first, we have to draw the diagram where the two points A and B situated at a distance 9m and 4m respectively from the base of the tower C. CD is the height of the tower and \[\theta \] is the angle of elevation of the top the tower. Then we must find out the trigonometric ratios, \[\tan \theta \]and\[\tan ({90^ \circ } - \theta )\]. Then comparing the ratios we can get CD which is the height of the tower.

Complete step by step answer:
Trigonometric ratios or trigonometric identities relate the lengths of the sides of a right angled triangle to its interior angles. There are six trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec) and cosecant (cosec). For a specific angle\[\theta \], the trigonometric ratios can be expressed in terms of sides of a right angled triangle. For a given right angled\[\vartriangle ABC\]. The ratios are given as follows.
\[\begin{gathered}
  \cos \theta = \dfrac{b}{h} \\
  \tan \theta = \dfrac{p}{b} \\
  \cot \theta = \dfrac{b}{p} \\
  \sec \theta = \dfrac{h}{b} \\
  \cos ec\theta = \dfrac{h}{p} \\
\end{gathered} \]

From the figure we get that \[AC = 9m\] and \[BC = 4m\].
We know that the trigonometric ratio, tangent of an angle is given by
\[\tan \theta = \dfrac{p}{b}\] …………………………………… (1)
Now in\[\vartriangle ACD\],
\[\tan \theta = \dfrac{{CD}}{{AC}} = \dfrac{{CD}}{9}\] …………………………………. (2)
And in\[\vartriangle BCD\],
\[\begin{gathered}
   \Rightarrow \tan ({90^ \circ } - \theta ) = \dfrac{{CD}}{{BC}} \\
   \Rightarrow \cot \theta = \dfrac{{CD}}{4} \\
\end{gathered} \]
……………………………………… (3)
We know the formula that
\[\cot \theta = \dfrac{1}{{\tan \theta }}\] ………………………………………… (4)
Now substituting the value of eq. (4) in eq. (3) we get
\[\begin{gathered}
   \Rightarrow \dfrac{1}{{\tan \theta }} = \dfrac{{CD}}{4} \\
   \Rightarrow \tan \theta = \dfrac{4}{{CD}} \\
\end{gathered} \]
…………………………………… (5)
Now comparing eq. (5) and (2), we will get,
\[\begin{gathered}
   \Rightarrow \dfrac{{CD}}{9} = \dfrac{4}{{CD}} \\
   \Rightarrow C{D^2} = 4 \times 9 \\
   \Rightarrow C{D^2} = 36 \\
   \Rightarrow CD = 6 \\
    \\
\end{gathered} \]
……………………………….. (6)
Hence it is proved that the height of the tower is 6m.
Note: The angle of elevation is defined as the angle from the horizontal upward to an object. The observer’s line of sight would be above the horizontal from the point of elevation.