The diagram shows a field $$\(A B C D\)$$. The bearing of $$\(B\)$$ from $$\(A\)$$ is $$\(140^{\circ}\)$$. $$\(C\)$$ is due east of $$\(B\)$$ and $$\(D\)$$ is due north of $$\(C\)$$. $$\(A B=400 \mathrm{~m}, B C=350 \mathrm{~m}\)$$ and $$\(C D=450 \mathrm{~m}\)$$. Find the bearing of $$\(D\)$$ from $$\(B\)$$. .............................................
Exam No：0580_s20_qp_42 Year：2020 Question No：5(a)
Answer:
\([0] 38\) or \([0] 37.9\) or \([0] 37.87 \ldots\)
Knowledge points:
E4.7.3 angles formed within parallel lines
E6.1 Interpret and use three-figure bearings. (Measured clockwise from the North, i.e. 000°–360°.)
E6.2.1 Apply Pythagoras’ theorem and the sine, cosine and tangent ratios for acute angles to the calculation of a side or of an angle of a right- angled triangle. (Angles will be quoted in degrees. Answers should be written in degrees and decimals to one decimal place.)
E6.2.2 Solve trigonometric problems in two dimensions involving angles of elevation and depression.
E6.2.3 Know that the perpendicular distance from a point to a line is the shortest distance to the line.
Solution:
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