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**X is due east point of y on a coast. Z is another point on the coast but 6.0km due south of Y. If the distance ZX is 12km, calculate the bearing of Z from X**

**A.**** 240°**

**B.**** 150°**

**C.**** 60°**

**D.**** 270°**

**QUICK ANSWER…**

**A**

**DETAILS…**

**If ****X is due east point of ****Y**** on a coast****,**** Z is another point on the coast but 6.0km due south of Y****, and ****the distance ZX is 12km, the bearing of Z from X**** will be = 240ᵒ. Let’s carefully resolve this…**

**First, lets sketch the problem as follows:**

**From the sketch, the bearing of Z from X is indicated by the arrow which will be = 270ᵒ – φ or 360ᵒ – 90 – φ**

**Solving for φ, we use SOHCAHTOA**

**Thus,**

**Sine φ = opp/hyp = 6/12 = 0.5**

**Φ = sin¯¹0.5 = 30ᵒ**

**Thus, the bearing of Z from X is = 270 – 30 = 240ᵒ.**

**Now for the right answer to the above question:**

**Option A is correct.****Option B is not correct.****C is incorrect.****D is not the correct answer.**

**KEY-POINTS…**

**You may please note these/this:**

**The bearing of a point from another point is an angle formed when you draw a clockwise arrow from the north pole of the reference point to the line linking the reference point to the desired point.****Please observe that in the problem above, the bearing arrow was drawn from the north of X to line ZX.**

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**/ culled from 2021 JAMB-UTME MATHEMATICS question 38 /**