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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

136°−46°=90°⇒cos136°=−sin46°;sin136°=cos46°v1=(20∠136°)=(20cos136°20sin136°)=(−20sin46°20cos46°)v2=(5∠46°)=(5cos46°5sin46°)v=v1+v2=5(−4sin46°+cos46°4cos46°+sin46°)direction=angle(v)==180°+tan−14cos46°+sin46°−4sin46°+cos46°≈122.96°distance=3∣v∣==15(−4sin46°+cos46°)2+(4cos46°+sin46°)2=(−4s+c)2+(4c+s)2==(16s2−8sc+c2)+(s2+8sc+16c2)==17s2+17c2=17(s2+c2)=1517≈61.85km

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