Solution:-
p+q = r or p-r = -q
(p-r).(p-r) = (-q).(-q)
p2+r2-2prcosθ1=q2
cosθ1 = 1/2 (as p=q=r)
θ1 = 60degree
when p+q+r = 0 or p+r = -q
(p+r).(p+r) = (-q)(-q)
p2+r2+2prcosθ2 = q2
cosθ2 = -1/2 (as p=q=r)
θ2 = 120 degree
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Hello aspirant,
Given P+Q=R and P=Q=R
When P+Q=R or P - R = - Q
(P - R) (P - R) = (- Q)(- Q)
P2-R2-2PR cos o1°(teta)=Q2
Coso1° = 1/2(as P=Q=R)
o1°=60°
Therefore,angle between Pand Q is 60°
I think my suggestions and knowledge may be helpful to us to u.
Thank you!!!
This may help you ...
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