Question

a boomerang is thrown with the following displacements: 4.5-m to the east, 7.5-m 30° north of west, and 3.0-m 10° south of west. what is the resultant displacement?

Answer 1

Solution.

$AB=4.5m;$

$BC=7.5m;$

$CD=3m;$

$\angle B=30^o;$

$\angle DCB=10^o;$

$AC^2=AB^2+BC^2-2ABBCcos\angle B;$

$AC^2=4.5^2+7.5^2-2\sdot4.5\sdot7.5\sdot 0.866=18.045;$

$AC=4.25m;$

$\dfrac{AC}{sin\angle B}=\dfrac{AB}{sin \angle C}\implies sin\angle C=\dfrac{ABsin\angle B}{AC}$ ;

$sin\angle C=\dfrac{4.5\sdot 0.5}{4.25}=0.529;$

$\angle C=32^o;$

$\angle ACD=22^o;$

$AD^2=AC^2+CD^2-2ACCDcos\angle ACD;$

$AD^2=4.25^2+3^2-2\sdot4.25\sdot3\sdot0.9272=3.4189;$

$AD=1.85m;$

$\dfrac{AD}{sin\angle ACD}=\dfrac{CD}{sin \angle CAD}\implies sin\angle CAD=\dfrac{CDsin\angle ACD}{AD};$

$sin\angle CAD=\dfrac{3\sdot 0.3746}{1.85}=0.6074;$

$\angle CAD=37^o;$

$\angle BAD=118^o-37^o=81^o;$

Answer: $AD=1.85m, \angle BAD=81^o;$ ,