Question

The angles of elevation of a cliff at a point A on the ground and a point B 100 m vertically above are \[60^\circ \] and \[30^\circ \] respectively. The height of cliff is
A. \[100\sqrt 3 {\text{ m}}\]
B. \[150\sqrt 3 {\text{ m}}\]
C. \[150{\text{ m}}\]
D. \[\dfrac{{200}}{{\sqrt 3 }}{\text{ m}}\]

Answer 1

Hint: Here, we will first draw the triangle using the given condition to simplify the calculation. Then use the tangential property, that is, \[\tan {\text{A}} = \dfrac{p}{b}\], where \[p\] is the perpendicular and \[b\] is the base. Apply this property, and then use the given conditions to find the required value.

Complete step by step answer
Given that the angles of elevation of a chiff at a point A on the ground and a point B 100 m vertically above are \[60^\circ \] and \[30^\circ \] respectively.

Let us assume that the height of the tower DC is \[h\] and the length AC and BE is \[x\].

First, we will draw the triangle using the given conditions.



We will first take the triangle \[\Delta {\text{ACD}}\],

We will use the tangential property, that is, \[\tan {\text{A}} = \dfrac{p}{b}\], where \[p\] is the perpendicular and \[b\] is the base.

Using the above tangential property, we get

\[\tan 60^\circ = \dfrac{{{\text{DC}}}}{{{\text{AC}}}}\]

Substituting the values of the length AC and DC in the above equation, we get

\[
  \tan 60^\circ = \dfrac{h}{x} \\
  \sqrt 3 = \dfrac{h}{x} \\
 \]

Cross-multiplying the above equation, we get

\[h = \sqrt 3 x{\text{ ......}}\left( 1 \right)\]

We will now take the triangle \[\Delta {\text{BCD}}\],

We will use the tangential property, that is, \[\tan {\text{A}} = \dfrac{p}{b}\], where \[p\] is the perpendicular and \[b\] is the base.

Using the above tangential property, we get

\[\tan 30^\circ = \dfrac{{{\text{DC}}}}{{{\text{BE}}}}\]

Substituting the values of the length DE and BE in the above equation, we get

\[
  \tan 30^\circ = \dfrac{{h - 100}}{x} \\
  \dfrac{1}{{\sqrt 3 }} = \dfrac{{h - 100}}{x} \\
 \]

Cross-multiplying the above equation, we get\[150\]

\[\sqrt 3 \left( {h - 100} \right) = x{\text{ ......}}\left( 2 \right)\]

From equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\], we get

\[
   \Rightarrow \sqrt 3 \left( {\sqrt 3 \left( {h - 100} \right)} \right) = h \\
   \Rightarrow \sqrt 3 \left( {\sqrt 3 h - 100\sqrt 3 } \right) = h \\
   \Rightarrow 3h - 300 = h \\
   \Rightarrow 3h - h = 300 \\
   \Rightarrow 2h = 300 \\
   \Rightarrow h = \dfrac{{300}}{2} \\
   \Rightarrow h = 150{\text{ m}} \\
 \]

Thus, the height of the tower is 150 meters.

Note: In solving these types of questions, you should be familiar with the concept of angle of elevation and the tangential properties. Students should not use any other trigonometric functions like sine or cosine of the given angle because the given condition of the problem, which is provided to us, is of the height of the triangle and a base. So, finding the hypotenuse for the sine and cosine function will be time taking to find the required value.