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?
1 answer:
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.
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