PLEASE HELP ME WITH PRE CAL!!

A drone is flying over a college campus. The drone spots the cafeteria on the west end of campus at a 27 ° angle of depression and the sports complex on the east end of campus at a 34° angle of depression. If the cafeteria and sports complex are separated by a straight road 2.3 mi long, how high is the drone?

0.67 mi

Step-by-step explanation:

The diagram illustrating the question is shown in the attach photo.

In triangle DCA,

Opposite = H

Adjacent = b

Angel θ = 27°

Tan θ = Opp /Adj

Tan 27° = H/b

Cross multiply

H = b x Tan 27°... (1)

From triangle DSA,

The diagram illustrating the question is shown in attach photo.

In triangle DCA,

Opposite = H

Adjacent = 2.3 – b

Angel θ = 34°

Tan θ = Opp /Adj

Tan 34° = H/ 2.3 – b

Cross multiply

H = Tan 34° (2.3 – b) .. (2)

Equating equation (1) and (2)

b x Tan 27° = Tan 34° (2.3 – b)

0.5095b = 0.6745(2.3 – b)

0.5095b = 1.55135 – 0.6745b

Collect like terms

0.5095b + 0.6745b = 1.55135

1.184b = 1.55135

Divide both side by 1.184

b = 1.55135/1.184

b = 1.31 mi

Substitute the value of b into any of the equation to obtain the height (H). In this case we shall use equation 1.

H = b x Tan 27°

H = 1.31 x Tan 27°

H = 0.67 mi

Therefore, the height of the drone is 0.67 mi

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

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