Question

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Answer 1

R² =P² +Q² +2PQ cosα

Since their magnitude is same, let us consider their magnitude to be P

P² = P² +P²+2P² cosα  =2P² +2P² cosα

=2P² (1+cosα)

1+cosα=  $\frac{1}{2}$

⇒ 2cos² $\frac{\alpha}{2}$ = $\frac{1}{2}$

⇒ cos² $\frac{\alpha}{2}$ = $\frac{1}{4}$  

⇒ cos  $\frac{\alpha}{2}$= $\frac{1}{2}$

= cos 60

⇒   $\frac{\alpha}{2}$ = 60

⇒ α=120  

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Answer 2

Answer:

Two vectors (inclined at any angle) and their sum vector from a triangle.

It is given that two Vector have a resultant equal to either of them, hence these three Vector from an equilateral triangle each angle of 60∘60∘. In the figure A→A→ and B→B→ are two Vectors (A→=B→)(A→=B→) having their sum vectors R→R→ such that

R→=A→=B→R→=A→=B→

Thus, the vector A→A→ and B→B→ of same magnitudes have the resultant Vectors R→R→ of the same magnitude. In this case angle between A→A→ and B→B→ is 120∘120∘.