Question

Kayson is looking at two buildings, building A and building B, at an angle of elevation of 40°. Building A is 120 feet away, and building B is 60 feet away. Which building is taller and by approximately how many feet? Building A; it is around 100.69 feet taller than building B Building A; it is around 50.34 feet taller than building B Building B; it is around 100.69 feet taller than building A Building B; it is around 50.34 feet taller than building A

Answer 1

Building A is bigger then building B

Answer 2

Answer:

Building A is taller. it is around 50.34 feet taller than building B

Step-by-step explanation:

Refer the attached figure

We are given that Kayson is looking at two buildings,building A and building B, at an angle of elevation of 40°.

$\angle ABC= 40^{\circ}\\\angle DBE = 40^{\circ}$

Building A is 120 feet away.

BC =120 feet

Building B is 60 feet away.

BE= 60 feet

In ΔABC

$tan \theta = \frac{Perpendicular}{Base}\\Tan 40^{\circ}=\frac{AC}{BC}\\Tan 40^{\circ}=\frac{AC}{120}\\0.839 \times 120=AC\\100.68=AC$

In ΔDBE

$tan \theta = \frac{Perpendicular}{Base}\\Tan 40^{\circ}=\frac{DE}{BE}\\Tan 40^{\circ}=\frac{DE}{60}\\0.839 \times 60=DE$

50.34=DE

Difference between their heights=100.68-50.34=50.34

So, Building A is taller. it is around 50.34 feet taller than building B