Question

Andrew must find the height of the city building. He used his book, which has a right angle at each of its four edges, to
determine a right triangle from his line of site to the top and bottom of the building along the edges
of the book. He
was 15
feet away from the building and his eyes are 3ft from the ground. How tall is the building?
Round your answers to two decimal places.

Answer 1

Answer:

$78\; {\rm ft}$.

Step-by-step explanation:

The goal is to find the length of ${\rm WY}$ in the diagram. To do so, make use of the fact that $RT \triangle {\rm YZX}$ (right triangle ${\rm YZX}$) is similar to $RT \triangle {\rm YXW}$.

To see why these two triangles are similar, note that:

• Both triangles include the angle ${\angle {\rm XYW}$, which is not $90^{\circ}$ (a shared angle,) and
• Each triangle includes a right angle: $\angle {\rm YZX} = \angle {\rm YXW} = 90^{\circ}$ (another shared angle.)

Hence, $RT \triangle {\rm YZX} \sim RT \triangle {\rm YXW}$. The ratio between the corresponding sides  of the two triangles would be equal:

$\begin{aligned}\frac{({\rm YX})}{({\rm YZ})} = \frac{({\rm YW})}{({\rm YX})}\end{aligned}$.

The length of $({\rm YX})$ can be found using Pythagorean Theorem within $RT \triangle {\rm YZX}$:

$\begin{aligned}({\rm YX}) &= \sqrt{({\rm YZ})^{2} + ({\rm ZX})^{2}}\end{aligned}$.

Rearrange the equation of the ratios to find the length of side $({\rm YW})$:

$\begin{aligned}({\rm YW}) &= ({\rm YX})\, \frac{({\rm YX})}{({\rm YZ})} \\ &= \frac{({\rm YX})^{2}}{({\rm YZ})} \\ &= \frac{({\rm YZ})^{2} + ({\rm ZX})^{2}}{({\rm YZ})} \\ &= \frac{3^{2} + 15^{2}}{3} \\ &= 78 \end{aligned}$.