Question

A blimp is 1100 meters high in the air and measures the angles of depression to two stadiums to the west of the blimp. If those measurements are 75.2° and 17.9°, how far apart are the two stadiums?

Answer 1

tanα=height/distance

d=h/tanα

d1=1100/tan17.9

d2=1100/tan75.2

So the distance between them is:

d=d1-d2

d=1100/tan17.9-1100/tan75.2 meters

d≈3115.03 m  (to nearest hundredth of a meter or centimeter)

Answer 2

The angle of depression represents the angle from a horizontal layout to a lower surface. The distance between the two stadiums is 3115.1 meters

The given parameters have been illustrated using the attached image of triangles.

The stadiums are represented with A and B.

First, calculate distance BO using:

$\tan T =\frac{BO}{TO}$

Where:

$\angle T = 90 -75.2 = 14.8$

$TO = 1100$

So, we have:

$\tan(14.8^o) = \frac{BO}{1100}$

Make BO the subject

$BO = 1100 * \tan(14.8^o)$

$BO = 1100 * 0.2642$

$BO = 290.62$

Next, calculate distance AO using:

$\tan T =\frac{AO}{TO}$

But in this case:

$\angle T = 90 -17.9 = 72.1$

$TO = 1100$

So, we have:

$\tan(72.1^o) = \frac{AO}{1100}$

Make AO the subject

$AO = 1100 * \tan(72.1^o)$

$AO = 1100 * 3.0961$

$AO = 3405.71$

The distance AB between the 2 stadiums is:

$AB = AO - BO$

$AB = 3405.71-290.61$

$AB = 3115.1$

Hence, the distance between the 2 stadiums is 3115.1 meters.

Read more about angles of depression at:

https://brainly.com/question/13697260