Given that A- - B- - C- If -A- - 4- -B- - 5 and -C- - -61- the angle between A and B is

The correct option is B 60^o A⃗ +B⃗ =C⃗ ⇒ | C⃗ #160; | =√( A ^ 2 + B ^ 2 +2ABcosθ #160; ) Where #160; θ is angle b/w A and B . ...

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Byju's Answer

Standard XII

Mathematics

Addition of Vectors

Given that ...

Question

Given that $\to A + \to B = \to C$ . If $| \to A | = 4$ , $| \to B | = 5$ and $| \to C | = \sqrt{61}$ , the angle between $A$ and $B$ is

30^{o}

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60^{o}

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90^{o}

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120^{o}

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Solution

The correct option is B $60^{o}$
$\to A + \to B = \to C \Rightarrow ∣ ∣ \to C ∣ ∣ = \sqrt{A^{2} + B^{2} + 2 A B cos θ}$
Where $θ$ is angle b/w $A$ and $B$ .
$\Rightarrow 61 = 16 + 25 + 2 \times 4 \times 5$ $cos θ$
$\Rightarrow cos θ = \frac{20}{40} = \frac{1}{2} \Rightarrow θ = 60^{0}$

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Given that →A+→B=→C. If |→A|=4, |→B|=5 and |→C|=√61, the angle between A and B is

The correct option is B 60o→A+→B=→C⇒∣∣→C∣∣=√A2+B2+2ABcosθWhere θ is angle b/w A and B.⇒61=16+25+2×4×5 cosθ⇒cosθ=2040=12⇒θ=600

Given that $\to A + \to B = \to C$ . If $| \to A | = 4$ , $| \to B | = 5$ and $| \to C | = \sqrt{61}$ , the angle between $A$ and $B$ is

The correct option is B $60^{o}$
$\to A + \to B = \to C \Rightarrow ∣ ∣ \to C ∣ ∣ = \sqrt{A^{2} + B^{2} + 2 A B cos θ}$
Where $θ$ is angle b/w $A$ and $B$ .
$\Rightarrow 61 = 16 + 25 + 2 \times 4 \times 5$ $cos θ$
$\Rightarrow cos θ = \frac{20}{40} = \frac{1}{2} \Rightarrow θ = 60^{0}$