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3-1.Vectors
easy
Two forces of magnitude $8 \,N$ and $15 \,N$ respectively act at a point. If the resultant force is $17 \,N$, the angle between the forces has to be .......
A
$60$
B
$45$
C
$90$
D
$30$
Solution
(c)
$R=\sqrt{A^2+B^2+2 A B \cos \theta}$
$A=8, B=15, R=17$
$17^2=8^2+15^2+2 \times 8 \times 15 \times \cos \theta$
$289=64+225+240 \cos \theta$
$\Rightarrow 289=289+24 \cos \theta$
$24 \cos \theta=0$
$\cos \theta=0 \Rightarrow \theta=90^{\circ}$
Standard 11
Physics
