Question

Difficulty: Very hardAngles of Elevation, Depression, and Bearings

A search-and-rescue helicopter leaves a central station PP and flies 16 km16\text{ km} on a bearing of 050050^\circ to reach a waypoint QQ. It then changes course and flies 12 km12\text{ km} on a bearing of 140140^\circ to reach a mountain rescue site RR. From the central station PP, the angle of elevation to the helicopter hovering vertically above point RR is 4545^\circ. What is the vertical height of the helicopter above the horizontal plane of station PP, in kilometers?

Answer: 20 km

Answer

The vertical height of the helicopter above the horizontal plane of station PP is 20 km20\text{ km}.
The horizontal journey forms a right-angled triangle PQRPQR with side lengths 16 km16\text{ km} and 12 km12\text{ km}, yielding a hypotenuse (horizontal distance PRPR) of 20 km20\text{ km}. Since the angle of elevation from PP to the hovering helicopter is 4545^\circ, the vertical height is equal to 20×tan(45)=20 km20 \times \tan(45^\circ) = 20\text{ km}.

Step-by-Step Solution

1
Find the back bearing of station PP from waypoint QQ
Back bearing = 050+180=230050^\circ + 180^\circ = 230^\circ
To determine the enclosed interior angle at point QQ, the reverse direction from QQ to PP must be calculated.
2
Calculate the interior angle PQR\angle PQR
\angle PQR = 230^\circ - 140^\circ = 90^\circ
Subtracting the forward bearing of RR from the back bearing of PP yields the right angle between the two paths.
3
Calculate the horizontal displacement distance PRPR
PR = \sqrt{16^2 + 12^2} = 20\text{ km}
Since PQR\triangle PQR is right-angled at QQ, the distance PRPR is obtained using the Pythagorean theorem.
4
Determine the vertical altitude using trigonometry
\text{Height} = PR \times \tan(45^\circ) = 20 \times 1 = 20\text{ km}
In the vertical right-angled triangle formed by PP, the ground projection of RR, and the helicopter, tan(45)=HeightHorizontal Distance\tan(45^\circ) = \frac{\text{Height}}{\text{Horizontal Distance}}.

Key Concept

Combining 3-point bearings in 2D with right-triangle trigonometry for 3D angles of elevation.
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