1. ## Trig/Vector help!

I tried this question and managed to get an answer of 56.5km/h using the cosine law, however I can still not get a heading angle and am not sure if I used the formular correctly.

A bird flies at a speed of 22km/h at a bearing of 265degrees. If the wind is blowing at 50km/h in a bearing direction of 10degrees, determine the resultant velocity (side x + direction) to a nearest tenth and degree, respectively.

Need some help...thanks =]

2. Hello, rebeccasnarfer!

$\text{A bird flies at a speed of 22 km/hr at a bearing of 265}^o.$
$\text{If the wind is blowing at 50 km/hr in a bearing direction of 10}^o,$
$\text{determine the resultant speed to the nearest tenth of a kilometer,}$
$\text{and bearing to the nearest degree.}$
Code:
                C         N
*         :
P        *  *       :
:       *     *     :
:      *        *   :
:  50 *           * :
:    *              * A
:10d*           *85d:
:  *        *       :
: * 75d * 22        :
:*  *               :
B *                   :
S

The bird flies from $\,A$ to $\,B$ at 22 km/hr: . $ABV = 22.$

Major angle $NAB = 265^o \quad\Rightarrow\quad \angle BAS = 85^o.$

The wind blows from $\,B$ to $\,C$ at 50 km/hr: . $BC = 50.$

$\angle PBC = 10^o \quad\Rightarrow\quad \angle CBA = 75^o$

In $\Delta ABC$, use the Law of Cosines:

. . $AC^2 \;=\;50^2 + 22^2 - 2(50(22)\cos75^o \:=\:2414.598101$

Hence: . $AC \:\approx\:49.1\text{ km/hr.}$

Let $A \,=\,\angle CAB.$

Law of Cosines: . $\cos A \;=\;\dfrac{22^2 + 49.1^2 - 50^2}{2(22)(49.1)} \;=\;0.182748565$

. . Hence: . $\angle A \;=\;79.47010295^o \;\approx\;79^o$

Then: . $\angle CAN \;=\;180^o - 79^o - 85^o \;=\;16^o$

The bird is flying at $49.1$ km/hr at a bearing of $344^o.$