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Question Number 34231 by byaw last updated on 03/May/18

Answered by MJS last updated on 03/May/18

136°−46°=90° ⇒ cos 136°=−sin 46°; sin 136°=cos 46° v_1 =(20 ∠136°)= (((20cos 136°)),((20sin 136°)) )= (((−20sin 46°)),((20cos 46°)) ) v_2 =(5 ∠46°)= (((5cos 46°)),((5sin 46°)) ) v=v_1 +v_2 =5 (((−4sin 46°+cos 46°)),((4cos 46°+sin 46°)) ) direction=angle(v)= =180°+tan^(−1) ((4cos 46°+sin 46°)/(−4sin 46°+cos 46°))≈122.96° distance=3∣v∣= =15(√((−4sin 46°+cos 46°)^2 +(4cos 46°+sin 46°)^2 ))= (−4s+c)^2 +(4c+s)^2 = =(16s^2 −8sc+c^2 )+(s^2 +8sc+16c^2 )= =17s^2 +17c^2 =17(s^2 +c^2 ) =15(√(17))≈61.85km

$$\mathrm{136}°−\mathrm{46}°=\mathrm{90}°\:\Rightarrow\:\mathrm{cos}\:\mathrm{136}°=−\mathrm{sin}\:\mathrm{46}°;\:\mathrm{sin}\:\mathrm{136}°=\mathrm{cos}\:\mathrm{46}° \\ $$$${v}_{\mathrm{1}} =\left(\mathrm{20}\:\angle\mathrm{136}°\right)=\begin{pmatrix}{\mathrm{20cos}\:\mathrm{136}°}\\{\mathrm{20sin}\:\mathrm{136}°}\end{pmatrix}=\begin{pmatrix}{−\mathrm{20sin}\:\mathrm{46}°}\\{\mathrm{20cos}\:\mathrm{46}°}\end{pmatrix} \\ $$$${v}_{\mathrm{2}} =\left(\mathrm{5}\:\angle\mathrm{46}°\right)=\begin{pmatrix}{\mathrm{5cos}\:\mathrm{46}°}\\{\mathrm{5sin}\:\mathrm{46}°}\end{pmatrix} \\ $$$${v}={v}_{\mathrm{1}} +{v}_{\mathrm{2}} =\mathrm{5}\begin{pmatrix}{−\mathrm{4sin}\:\mathrm{46}°+\mathrm{cos}\:\mathrm{46}°}\\{\mathrm{4cos}\:\mathrm{46}°+\mathrm{sin}\:\mathrm{46}°}\end{pmatrix} \\ $$$$\mathrm{direction}=\mathrm{angle}\left({v}\right)= \\ $$$$=\mathrm{180}°+\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{4cos}\:\mathrm{46}°+\mathrm{sin}\:\mathrm{46}°}{−\mathrm{4sin}\:\mathrm{46}°+\mathrm{cos}\:\mathrm{46}°}\approx\mathrm{122}.\mathrm{96}° \\ $$$$\mathrm{distance}=\mathrm{3}\mid{v}\mid= \\ $$$$=\mathrm{15}\sqrt{\left(−\mathrm{4sin}\:\mathrm{46}°+\mathrm{cos}\:\mathrm{46}°\right)^{\mathrm{2}} +\left(\mathrm{4cos}\:\mathrm{46}°+\mathrm{sin}\:\mathrm{46}°\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(−\mathrm{4}{s}+{c}\right)^{\mathrm{2}} +\left(\mathrm{4}{c}+{s}\right)^{\mathrm{2}} = \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{16}{s}^{\mathrm{2}} −\mathrm{8}{sc}+{c}^{\mathrm{2}} \right)+\left({s}^{\mathrm{2}} +\mathrm{8}{sc}+\mathrm{16}{c}^{\mathrm{2}} \right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{17}{s}^{\mathrm{2}} +\mathrm{17}{c}^{\mathrm{2}} =\mathrm{17}\left({s}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$=\mathrm{15}\sqrt{\mathrm{17}}\approx\mathrm{61}.\mathrm{85}{km} \\ $$

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