JAMB 2018 Mathematics · Question 33

From a point R, 300m north of P, a man walks eastwards to a place; Q which is 600m from P. Find the bearing of P from Q correct to the nearest degree

Options

  1. A.026o^o
  2. B.045o^o
  3. C.210o^o
  4. D.240o^o

Answer: D. 240o^o

Explanation written with AI and checked against the answer key

1. Analyze the question

We are given two points, P and R. Point R is 300m North of P. A man walks East from R to a point Q, such that the distance from P to Q is 600m. We need to find the bearing of P from Q.

2. Explain the concept
  • Geometry: We have a right-angled triangle △PRQ\triangle PRQ, where ∠PRQ=90∘\angle PRQ = 90^\circ (because R is North of P and Q is East of R).
  • Trigonometry: In a right-angled triangle, we can use SOH-CAH-TOA.
  • Bearing: The bearing of P from Q is the angle measured clockwise from the North line at Q to the line QP.
3. Show the solution

Step 1: Visualize the triangle

  • PR=300mPR = 300\text{m} (Vertical side)
  • PQ=600mPQ = 600\text{m} (Hypotenuse)
  • ∠RPQ\angle RPQ is the angle at P inside the triangle.

Using trigonometry in △PRQ\triangle PRQ: cos⁡(∠RPQ)=AdjacentHypotenuse=PRPQ\cos(\angle RPQ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PR}{PQ} cos⁡(∠RPQ)=300600=0.5\cos(\angle RPQ) = \frac{300}{600} = 0.5 ∠RPQ=cos⁡−1(0.5)=60∘\angle RPQ = \cos^{-1}(0.5) = 60^\circ

Step 2: Determine the bearing

  • At point Q, draw a North line. Since Q is East of R, the line RQ is horizontal (West-East).
  • The angle of the line QP relative to the South line at Q is equal to the alternate angle ∠RPQ=60∘\angle RPQ = 60^\circ.
  • The bearing is measured clockwise from the North line at Q.
  • The angle from North to South is 180∘180^\circ.
  • The bearing of P from Q is 180∘+(angle between South and QP)180^\circ + (\text{angle between South and QP}).
  • Since the angle between the vertical (South) and QP is 60∘60^\circ, the bearing is 180∘+60∘=240∘180^\circ + 60^\circ = 240^\circ.
4. Identify the answer

The calculated bearing is 240∘240^\circ. Looking at the options provided: A. 026∘026^\circ B. 045∘045^\circ C. 210∘210^\circ D. 240∘240^\circ

The correct option is D.

5. Tip

When dealing with bearing questions, always draw a neat sketch. Remember that "North" is always 000∘/360∘000^\circ/360^\circ, "East" is 090∘090^\circ, "South" is 180∘180^\circ, and "West" is 270∘270^\circ. Using alternate angles (the "Z" rule) is the fastest way to transfer angles between your North lines!

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