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A-boy-starts-from-a-point-A-and-moves-on-a-bearing-of-20-to-a-point-B-which-is-5km-from-A-He-then-changes-his-course-to-a-bearing-of-11-and-moves-to-a-point-C-which-is-12km-from-B-Find-the-distanc

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Question Number 40055 by Rio Mike last updated on 15/Jul/18
A boy starts from a point A and moves on a bearing of 20° to a point B which is 5km from A.He then changes his course to a bearing of 11° and moves to a point C which is 12km from B. Find the distance and bearing from C to A.
$${A}\:{boy}\:{starts}\:{from}\:{a}\:{point}\:{A} \\ $$$${and}\:{moves}\:{on}\:{a}\:{bearing}\:{of} \\ $$$$\mathrm{20}°\:{to}\:{a}\:{point}\:{B}\:{which}\:{is} \\ $$$$\mathrm{5}{km}\:{from}\:{A}.{He}\:{then}\:{changes} \\ $$$${his}\:{course}\:{to}\:{a}\:{bearing}\:{of}\: \\ $$$$\mathrm{11}°\:{and}\:{moves}\:{to}\:{a}\:{point}\:{C}\:{which}\:{is} \\ $$$$\mathrm{12}{km}\:{from}\:{B}. \\ $$$${Find}\:{the}\:{distance}\:{and}\:{bearing} \\ $$$${from}\:{C}\:{to}\:{A}. \\ $$$$ \\ $$
Answered by MJS last updated on 16/Jul/18
A= ((0),(0) ) B= (((5cos 20°)),((5sin 20°)) ) C=B+ (((12cos 11°)),((12sin 11°)) )= (((5cos 20° +12cos 11°)),((5sin 20° +12sin 11°)) ) ∣AC∣=(√((5cos 20° +12cos 11°)^2 +(5sin 20° +12sin 11°)^2 ))= =(√(25cos^2 20° +120cos 20° cos 11° +144cos^2 11° +25sin^2 20° +120sin 20° sin 11° +144sin^2 11°))= =(√(169+120(cos 20° cos 11° +sin 20° sin 11°)))= =(√(169+120(((cos(20°−11°)+cos(20°+11°))/2)+((cos(20°−11°)−cos(20°+11°))/2))))= =(√(169+120cos 9°))≈16.956
$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${B}=\begin{pmatrix}{\mathrm{5cos}\:\mathrm{20}°}\\{\mathrm{5sin}\:\mathrm{20}°}\end{pmatrix} \\ $$$${C}={B}+\begin{pmatrix}{\mathrm{12cos}\:\mathrm{11}°}\\{\mathrm{12sin}\:\mathrm{11}°}\end{pmatrix}=\begin{pmatrix}{\mathrm{5cos}\:\mathrm{20}°\:+\mathrm{12cos}\:\mathrm{11}°}\\{\mathrm{5sin}\:\mathrm{20}°\:+\mathrm{12sin}\:\mathrm{11}°}\end{pmatrix} \\ $$$$\mid{AC}\mid=\sqrt{\left(\mathrm{5cos}\:\mathrm{20}°\:+\mathrm{12cos}\:\mathrm{11}°\right)^{\mathrm{2}} +\left(\mathrm{5sin}\:\mathrm{20}°\:+\mathrm{12sin}\:\mathrm{11}°\right)^{\mathrm{2}} }= \\ $$$$=\sqrt{\mathrm{25cos}^{\mathrm{2}} \:\mathrm{20}°\:+\mathrm{120cos}\:\mathrm{20}°\:\mathrm{cos}\:\mathrm{11}°\:+\mathrm{144cos}^{\mathrm{2}} \:\mathrm{11}°\:+\mathrm{25sin}^{\mathrm{2}} \:\mathrm{20}°\:+\mathrm{120sin}\:\mathrm{20}°\:\mathrm{sin}\:\mathrm{11}°\:+\mathrm{144sin}^{\mathrm{2}} \:\mathrm{11}°}= \\ $$$$=\sqrt{\mathrm{169}+\mathrm{120}\left(\mathrm{cos}\:\mathrm{20}°\:\mathrm{cos}\:\mathrm{11}°\:+\mathrm{sin}\:\mathrm{20}°\:\mathrm{sin}\:\mathrm{11}°\right)}= \\ $$$$=\sqrt{\mathrm{169}+\mathrm{120}\left(\frac{\mathrm{cos}\left(\mathrm{20}°−\mathrm{11}°\right)+\mathrm{cos}\left(\mathrm{20}°+\mathrm{11}°\right)}{\mathrm{2}}+\frac{\mathrm{cos}\left(\mathrm{20}°−\mathrm{11}°\right)−\mathrm{cos}\left(\mathrm{20}°+\mathrm{11}°\right)}{\mathrm{2}}\right)}= \\ $$$$=\sqrt{\mathrm{169}+\mathrm{120cos}\:\mathrm{9}°}\approx\mathrm{16}.\mathrm{956} \\ $$

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