Question : Given that A+B=C if |A|=4, |B|=5 and |C|=√61, the angle between A and B in degree is 30n. Find n?
Doubt by Suhani
Solution :
A+B=C (Given)
A = 4
B = 5
C = √61
θ = 30n
A+B=C
(A+B).(A+B)=C.C
A.A+A.B+B.A+B.B=C.C
A²+A.B+B.A+B²=C² [∵A.A=A²]
A²+A.B+A.B+B²=C² [∵ A.B=B.A]
A²+2A.B+B²=C²
A²+2ABcosθ+B²=C² [∵ A.B=ABcosθ]
2ABcosθ=C²-A²-B²
cosθ=[C²-A²-B²]/2AB
cosθ=[(√61)²-(4)²-(5)²]/2(4)(5)
cosθ=[61-16-25]/40
cosθ=20/40
cosθ=1/2
cosθ=cos60
cos30n=cos60
30n=60
n=60/30
n=2
Hence, the required value of n is 2.
Similar Question :
1.) Given that A+B=C if |A|=4, |B|=5 and |C|=√61, the angle between A and B is
(a) 30°
(b) 60°
(c) 90°
(d) 120°
2.) If vectors A, B and C have magnitudes 8, 15 and 17 units and A+B=C, find the angle between A and B?
Click Here to see the Answer
1.) (b) 60°
2.) 90°