Question

A plane flying at a certain altitude is observed from two points that are 3 miles apart. The angles of elevation made by the two points are 55° and 72°, as seen in the diagram. What is the altitude of the plane to the nearest tenth?

Answer 1

Answer:

8.0 mi  

Explanation:

Let's call your triangle ABC as in the diagram. We can use the SAS property of triangles and the Law of Sines.

1. Determine m∠B

m∠ B + 72° = 180°

         m∠B = 108°

2. Determine b

By the Law of Sines,

b/sinB = c/sinC

       b = c(sinB/sinC)

          = 3 × (sin108°/sin17°)

          = 3 × 0.951/0.292

          = 9.76 mi

3. Calculate the area of ∆ABC

Area = ½ bc sinA = ½ × 9.76 × 3sin55° = ½ × 9.76 × 3 × 0.819 = 12.0 mi²

4. Calculate h

Area = ½ × base × height

12.0 mi² = ½ × 3 mi × h

h = 2 × 12.0 mi/3= 8.0 mi  

The altitude of the plane is 8.0 mi.