Question

A man of height 6 ft. observes the top of a tower and the foot of the tower at angles of \[45^\circ \] and \[30^\circ \] of elevation and depression respectively. Assuming the man to stand on the level ground, the height of the tower is:
 A) \[6\left( {\sqrt 3 - 1} \right)\] ft.
B) \[6\left( {\sqrt 3 + 1} \right)\] ft.
C) \[6\left( {\sqrt 2 - 1} \right)\] ft.
D) None of these

Answer 1

Hint: Use trigonometry ratio \[\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}\] in the triangle making \[30^\circ \] and find the base from this then use this trigonometry ratio in the triangle making \[45^\circ \] and use the obtained base value into this.

Complete Step-by-Step solution:
First, we will draw the figure,

Angle DEB is given in the question.

As we know that alternate interior angles are equal, so angle BCE will be the same as angle DEB that is .

In the triangle BAE, angle E is known and the perpendicular AB is known and we want to compute base AE so, we can use \[\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}\].

Now we are applying this trigonometric ratio into the triangle BAE and substituting the values into trigonometric ratio,

\[
  \tan 30^\circ = \dfrac{{AB}}{{EA}} \\
  \tan 30^\circ = \dfrac{6}{{EA}} \\
\]

As we know \[\tan 30^\circ = \dfrac{{\text{1}}}{{\sqrt 3 }}\] substituting this value and solve for EA,

\[
  \dfrac{1}{{\sqrt 3 }} = \dfrac{6}{{EA}} \\
  EA = 6\sqrt 3 \\
\]

Side EA is equivalent to side DB.

Thus, \[DB = EA = 6\sqrt 3 \]

Now we are applying trigonometric ratio \[\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}\] into the triangle DBC and substituting the values into trigonometric ratio,

\[\tan 45^\circ = \dfrac{{DC}}{{DB}}\]

As we know \[\tan 45^\circ = 1\] and we have \[DB = EA = 6\sqrt 3 \], substituting these values and solving,

\[
  1 = \dfrac{{DC}}{{6\sqrt 3 }} \\
  DC = 6\sqrt 3 \\
\]

From the figure, it is seen that the height of the building is the sum of ED and DC.

Now we have the value of ED and DC. So, we will add these to find the height of the building,

\[
  ED + DC = 6 + 6\sqrt 3 \\
   = 6\left( {1 + \sqrt 3 } \right) \\
\]

Thus, the height of the building is \[6\left( {1 + \sqrt 3 } \right)\] ft.

Hence option B is correct.

Note:
Any of the trigonometric functions can be used to find the answer but while solving these types of questions we have to first focus on “what we have to find” and then on “what data we have” so make our calculation shorter.