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Difficulty: MediumLaw of Sines and Law of Cosines

A drone is positioned in the sky above a flat field between two ground observation stations, Station AA and Station BB, which are 500500 meters apart. From Station AA, the angle of elevation to the drone is 40∘40^\circ. From Station BB, looking back toward the drone, the angle of elevation is 65∘65^\circ. Assuming the drone and both observation stations lie in the same vertical plane, which of the following expressions represents the direct distance, in meters, from Station AA to the drone?

  1. A
    500sin⁡(40∘)sin⁡(75∘)\frac{500 \sin(40^\circ)}{\sin(75^\circ)}
  2. 500sin⁡(65∘)sin⁡(75∘)\frac{500 \sin(65^\circ)}{\sin(75^\circ)}Answer
  3. C
    500sin⁡(75∘)sin⁡(65∘)\frac{500 \sin(75^\circ)}{\sin(65^\circ)}
  4. D
    500sin⁡(65∘)sin⁡(40∘)\frac{500 \sin(65^\circ)}{\sin(40^\circ)}
  5. E
    500sin⁡(40∘)sin⁡(65∘)\frac{500 \sin(40^\circ)}{\sin(65^\circ)}

Answer

The correct expression is 500sin⁡(65∘)sin⁡(75∘)\frac{500 \sin(65^\circ)}{\sin(75^\circ)}.
To find the distance from Station AA to the drone (ADAD), model the scenario as triangle ABDABD. The ground side AB=500AB = 500 meters. The interior angle at AA is 40∘40^\circ and at BB is 65∘65^\circ. The third angle at the drone DD is 180∘−40∘−65∘=75∘180^\circ - 40^\circ - 65^\circ = 75^\circ. By the Law of Sines, ADsin⁡(65∘)=500sin⁡(75∘)\frac{AD}{\sin(65^\circ)} = \frac{500}{\sin(75^\circ)}, which simplifies to AD=500sin⁡(65∘)sin⁡(75∘)AD = \frac{500 \sin(65^\circ)}{\sin(75^\circ)}.

Step-by-Step Solution

1
Determine the interior angle of the triangle at the drone's location (DD).
∠D=180∘−(40∘+65∘)=75∘\angle D = 180^\circ - (40^\circ + 65^\circ) = 75^\circ
The interior angles of any triangle must sum to 180∘180^\circ.
2
Apply the Law of Sines to relate the known side AB=500AB = 500 meters and its opposite angle ∠D=75∘\angle D = 75^\circ to the unknown side ADAD and its opposite angle ∠B=65∘\angle B = 65^\circ.
ADsin⁡(65∘)=500sin⁡(75∘)\frac{AD}{\sin(65^\circ)} = \frac{500}{\sin(75^\circ)}
The Law of Sines states that in any triangle, asin⁡(A)=bsin⁡(B)=csin⁡(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}.
3
Solve for the side length ADAD.
AD=500sin⁡(65∘)sin⁡(75∘)AD = \frac{500 \sin(65^\circ)}{\sin(75^\circ)}
Multiplying both sides of the equation by sin⁡(65∘)\sin(65^\circ) isolates ADAD.

Key Concept

Applying the Law of Sines to find an unknown side length given two angles and one side of a triangle.
Estimated Time:1m 15s
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Topics

Law of Sines and Law of Cosines

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