Question

Angle between two vectors of magnitudes $12$ and $18$ units, when their resultant is $24$ units is

• A. $82^{\circ} 31^{'}$
• B. $63^{\circ} 51^{'}$
• C. $89^{\circ} 16^{'}$
• D. $75^{\circ} 52^{'}$.

Answer 1

Answer: (D)

Given: Magnitude of first vector $(\vec{A})=12$;
Magnitude of second vector $(\vec{B})=18$
and resultant of the given vectors $(\vec{R})=24$.
We know that resultant vector $|\vec{R}|=24=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}$
$=\sqrt{(12)^{2}+(18)^{2}+2 \times 12 \times 18 \cos }$
or $(24)^{2}=144+324+432 \cos \theta$
$\Rightarrow 432 \cos \theta=108$
$\Rightarrow \cos \theta=\frac{108}{432}=0 \times 25$
or $\theta =\cos ^{-1} 0 \times 25=75^{\circ} 52' .$