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Difficulty: MediumAngles of Elevation, Depression, and Bearings

A field surveyor starts at point PP and walks 10 km10\text{ km} due East to point QQ. From point QQ, she changes direction and walks 10 km10\text{ km} on a bearing of 210∘210^\circ to reach point RR. What is the bearing of point PP from point RR?

  1. 330∘330^\circAnswer
  2. B
    060∘060^\circ
  3. C
    150∘150^\circ
  4. D
    300∘300^\circ

Answer

The bearing of point PP from point RR is 330∘330^\circ.
Point QQ lies 10 km10\text{ km} East of PP (bearing 090∘090^\circ). From QQ, moving 10 km10\text{ km} on bearing 210∘210^\circ forms an interior angle of 60∘60^\circ with the line QPQP (which has bearing 270∘270^\circ). Because PQ=QR=10 kmPQ = QR = 10\text{ km} and the included angle is 60∘60^\circ, △PQR\triangle PQR is equilateral, making ∠QRP=60∘\angle QRP = 60^\circ. The back bearing from RR to QQ is 030∘030^\circ. Subtracting the 60∘60^\circ interior angle from 030∘030^\circ yields a bearing of 330∘330^\circ for point PP from point RR.

Step-by-Step Solution

1
Determine the interior angle ∠PQR\angle PQR at point QQ.
∠PQR=60∘\angle PQR = 60^\circ
Since PP is due West of QQ, the bearing of PP from QQ is 270∘270^\circ. The bearing of RR from QQ is 210∘210^\circ. The interior angle between these two lines is 270∘−210∘=60∘270^\circ - 210^\circ = 60^\circ.
2
Analyze the properties of triangle PQRPQR.
Triangle PQRPQR is an equilateral triangle with side lengths 10 km10\text{ km} and interior angles of 60∘60^\circ.
Given PQ=10 kmPQ = 10\text{ km} and QR=10 kmQR = 10\text{ km}, △PQR\triangle PQR is isosceles. Since the vertex angle ∠PQR=60∘\angle PQR = 60^\circ, the remaining two angles are also 60∘60^\circ each.
3
Calculate the back bearing of QQ from RR.
Bearing of QQ from RR is 030∘030^\circ.
The bearing of RR from QQ is 210∘210^\circ. The back bearing is 210∘−180∘=030∘210^\circ - 180^\circ = 030^\circ.
4
Compute the bearing of PP from RR.
Bearing of PP from RR is 330∘330^\circ.
From line RQRQ (bearing 030∘030^\circ), line RPRP lies 60∘60^\circ counter-clockwise (since ∠QRP=60∘\angle QRP = 60^\circ). Thus, 030∘−60∘=−30∘≡330∘030^\circ - 60^\circ = -30^\circ \equiv 330^\circ.

Key Concept

Bearings and Triangle Geometry
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Topics

Angles of Elevation, Depression, and Bearings

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