A drone is flying over a college campus. the drone spots the cafeteria on the west end of campus at a 27 degree angle of depression and the sports complex on the east end of campus at a 34 degree angle of depression. If the cafeteria and sports complex are sped aged by a straight road 2.3 miles long how high is the drone please help I don’t get this

The question is an illustration of angles of depressions and elevations

The drone is 0.67 miles high

The angles of depression are given as:

The length of the road is given as:

See attachment for the image of the scenario

Considering triangle ABM, we have the following tangent ratio

Make x the subject

Considering triangle CBM, we have the following tangent ratio

Make x the subject

Substitute in

Evaluate tan 27 and tan 34

Take reciprocals

Collect like terms

Divide

Approximate

Hence, the drone is 0.67 miles high

Read more about angles of depression at:

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0.67 miles

Step-by-step explanation:

in the attached diagram, the drone is at B. The cafeteria is at A and the sport complex at C. We want to determine how high up the drone is, i.e. |BMM

Using Trigonometry:

In Right Triangle ABM

Similarly, in Right Triangle BMC

Therefore:

Since h=yTan 27^\circ

h=1.3102Tan 27^\circ

h=0.6676 miles

The height of the drone therefore is 0.67 miles (correct to 2 decimal places)

a precentage of 6th graders. the population is all 6th graders. let me know if i'm wrong, i hope this ! : )

step-by-step explanation: