Question

A surveyor stands at a window on the 9th floor of an office tower. He measures the angles of elevation and depression of the top and the base of a taller building. The surveyor sketches this plan of his measurements. Determine the height of the taller building to the nearest tenth of a meter.

Answer 1

Given,

The height of the building upto 9th floor is 39 meters.

The angle of elevation is 31 degree.

The angle of depression is 42 degree.

The diagram of the building and taller building is,

Consider,

AB is the height of the building upto 9th floor.

CE is the height of the taller biulding.

BE=AD is the distance between building and taller building.

Taking triangle ADE,

$\begin{gathered} \tan 42^{\circ}=\frac{DE}{AD} \\ \text{here, DE=AB=39m} \\ \tan 42^{\circ}=\frac{39}{AD} \\ AD=\frac{39}{\tan42^{\circ}} \\ =\frac{39}{0.900} \\ =43.34\text{ m} \end{gathered}$

The distance between the building and the taller building is 43.34 m (approximately).

Taking triangle ADC,

$\begin{gathered} \tan 31^{\circ}=\frac{CD}{AD} \\ \tan 31^{\circ}=\frac{CD}{43.34} \\ 0.601\times43.34=CD \\ CD=26.04734\text{ m} \\ \approx26.05\text{m} \end{gathered}$

The height of the taller building is,

$\text{Height of building =CD+DE=39+26.05=}65.05\approx65.1\text{ m}$

The height of the taller building is 65.1 meter.