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

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

[tex]Tan 27^\circ=\frac{h}{y} \\h=yTan 27^\circ[/tex]

Similarly, in Right Triangle BMC

[tex]Tan 34^\circ=\frac{h}{2.3-y} \\h=(2.3-y)Tan 34^\circ[/tex]

Therefore:

[tex]h=yTan 27^\circ =(2.3-y)Tan 34^\circ\\yTan 27^\circ =(2.3-y)Tan 34^\circ\\yTan 27^\circ =2.3Tan 34^\circ-yTan 34^\circ\\yTan 27^\circ +yTan 34^\circ=2.3Tan 34^\circ\\y(Tan 27^\circ+Tan 34^\circ)=2.3Tan 34^\circ\\y=\dfrac{2.3Tan 34^\circ}{Tan 27^\circ+Tan 34^\circ} \\y=1.3102[/tex]

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)

MrRoyal MrRoyal · Answer 2 · 2021-11-26T02:55:00+01:00

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:

[tex]\mathbf{\theta_1 = 27}[/tex]

[tex]\mathbf{\theta_2 = 34}[/tex]

The length of the road is given as:

[tex]\mathbf{l = 2.3\ miles}[/tex]

See attachment for the image of the scenario

Considering triangle ABM, we have the following tangent ratio

[tex]\mathbf{tan27 = \frac{h}{2.3 - x}}[/tex]

Make x the subject

[tex]\mathbf{x= 2.3 - \frac{h}{tan27 }}[/tex]

Considering triangle CBM, we have the following tangent ratio

[tex]\mathbf{tan34 = \frac{h}{ x}}[/tex]

Make x the subject

[tex]\mathbf{x= \frac{h}{tan34 }}[/tex]

Substitute [tex]\mathbf{x= 2.3 - \frac{h}{tan27 }}[/tex] in [tex]\mathbf{x= \frac{h}{tan34 }}[/tex]

[tex]\mathbf{2.3 - \frac{h}{tan27 }= \frac{h}{tan34 }}[/tex]

Evaluate tan 27 and tan 34

[tex]\mathbf{2.3 - \frac{h}{0.5095 }= \frac{h}{0.6745 }}[/tex]

Take reciprocals

[tex]\mathbf{2.3 - 1.9627h= 1.4826h }[/tex]

Collect like terms

[tex]\mathbf{1.4826h + 1.9627h= 2.3}[/tex]

[tex]\mathbf{3.4453h= 2.3}[/tex]

Divide

[tex]\mathbf{h= 0.6676}[/tex]

Approximate

[tex]\mathbf{h= 0.67}[/tex]

Hence, the drone is 0.67 miles high

Read more about angles of depression at:

https://brainly.com/question/13697260