Answer to Question #137244 in Trigonometry for Parkire

Question

A sailboat under auxiliary power is proceeding on a bearing 25° north of west at 6,25km/h in still water.Then a tail wind blowing 15km/h in the direction 35° south of west, alters the course of the sailboat.What is the resultant speed and direction of the saiboat (correct to 2 decimal places)?

Answer 1

From the above diagram, resultant velocity will be

$\vec v_r=\vec v_b+\vec v_w$

Thus,

$\vec v_r=(v_b\cos(25^0),v_b\sin(25^0))+(v_w\cos(35^0),-v_w\sin(35^0))\\ \vec v_r=(v_b\cos(25^0)+v_w\cos(35^0),v_b\sin(25^0)-v_w\sin(35^0))\\ \vec v_r=(578.73,255.53)\\ \implies v_r=\sqrt{578.73^2+255.53^2}\approx 632.63km/h$

Direction is,

$\phi=\tan^{-1}(\frac{255.53}{578.73})\approx23.82^0$

That is , direction is along $23.82^0$ north of west.