# Trig/Vector help!

• Nov 28th 2010, 07:05 PM
rebeccasnarfer
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 =]
• Nov 28th 2010, 08:34 PM
Soroban
Hello, rebeccasnarfer!

Quote:

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

                C        N                 *        :       P        *  *      :       :      *    *    :       :      *        *  :       :  50 *          * :       :    *              * A       :10d*          *85d:       :  *        *      :       : * 75d * 22        :       :*  *              :     B *                  :                           S

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

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

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

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

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

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

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

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

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

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

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

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