Chapter 14.3, Problem 51PPSE

Solution

The height of obelisk by setting trigonometric equations..

The height of obelisk is $61.73$ m.

Given data: The diagram of the obelisk of height y is shown in Fig 1. The distance between the angle of $\text{3}{\text{7}}^{\text{ο}}$ from the ground and top of the obelisk and an angle of $\text{3}{\text{0}}^{\text{ο}}$ from the ground to the top of the obelisk is 25 m.

  

Method/Formula used:

The right angle triangle $\Delta OBC$ the trigonometric ratio tangent is given by $\tan \theta =\frac{AB}{CB}$ .

  

Calculations:

Let the Obelisk RS of height y makes angle $\text{3}{\text{7}}^{\text{ο}}$ at point Q and $\text{3}{\text{0}}^{\text{ο}}$ at point P , also let distance QR equals x m. The distance between points P and Q is 25 m.

Therefore, in triangle QRS ,

  $\begin{array}{c}\tan \text{3}{\text{7}}^{\text{ο}}=\frac{y}{x}\\ x=\frac{y}{\tan \text{3}{\text{7}}^{\text{ο}}}\end{array}\}$ … (2)

Therefore, in triangle PRS ,

  $\begin{array}{c}\tan \text{3}{\text{0}}^{\text{ο}}=\frac{y}{x+25}\\ x+25=\frac{y}{\tan \text{3}{\text{0}}^{\text{ο}}}\end{array}\}$ … (3)

From results, (2) and (3),

  $\begin{array}{c}x+25=\frac{y}{\tan \text{3}{\text{0}}^{\text{ο}}}\\ x=\frac{y}{\tan \text{3}{\text{7}}^{\text{ο}}}\end{array}$

Subtract

  $\begin{array}{c}\left(x+25\right)-\left(x\right)=\frac{y}{\tan \text{3}{\text{0}}^{\text{ο}}}-\frac{y}{\tan \text{3}{\text{7}}^{\text{ο}}}\\ 25=y\left(\frac{1}{\tan \text{3}{\text{0}}^{\text{ο}}}-\frac{1}{\tan \text{3}{\text{7}}^{\text{ο}}}\right)\end{array}$

Substitute 1.732 for 1/ $\tan \text{3}{\text{0}}^{\text{ο}}$ and 1.327 for 1/ $\tan \text{3}{\text{7}}^{\text{ο}}$

  $\begin{array}{c}25=y\left(1.732-1.327\right)\\ y=\frac{25}{0.405}\\ y=61.73\text{}\end{array}$

Therefore, the height of obelisk is $61.73$ m.