Let $\to A and \to B$ be the two vectors of magnitude 10 unit each. If they are inclined to the $X - a x i s$ at angles $30^{\circ} and 60^{\circ}$ respectively, find the resultant.

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Solution

Consider the problem

$\to A$ and $\to B$ each of magnitudes same $= 10 u n i t$

Angle between $\to A$ and $\to B = 30^{\circ}$

Now,

Resultant,
$| \to R | = | \to A + \to B |$
then,
$R = \sqrt{A^{2} + B^{2} + 2 A B cos θ}$ (where $θ$ is angle between $\to A$ and $\to B$

$= \sqrt{100 + 100 + 2 (100) cos 30^{\circ}}$

$= 10 \sqrt{1 + 1 + \sqrt{3}}$

$= 10 \sqrt{2 + \sqrt{3}}$ unit.

Hence the resultant is $10 \sqrt{2 + \sqrt{3}}$ unit.

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Let →A and →B be the two vectors of magnitude 10 unit each. If they are inclined to the X−axis at angles 30∘ and 60∘ respectively, find the resultant.

Consider the problem →A and →B each of magnitudes same =10unitAngle between →A and →B=30∘ Now, Resultant,|→R|=|→A+→B|then, R=√A2+B2+2ABcosθ (where θ is angle between →A and →B =√100+100+2(100)cos30∘=10√1+1+√3=10√2+√3 unit.Hence the resultant is 10√2+√3 unit.

Let $\to A and \to B$ be the two vectors of magnitude 10 unit each. If they are inclined to the $X - a x i s$ at angles $30^{\circ} and 60^{\circ}$ respectively, find the resultant.