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Answer :

Answer:

the angle between u=(8, -2) and v=(9,3) is 32.5Β°

Step-by-step explanation:

u=(8,-2)=(u1,u2)β†’u1=8, u2=-2

v=(9,3)=(v1,v2)β†’v1=9, v2=3

We can find the angle between two vectors using the formula of dot product:

u . v =β•‘uβ•‘β•‘vβ•‘cos Ξ±   (1)

And the dot product is:

u . v = u1 v1 + u2 v2

u . v = (8)(9)+(-2)(3)

u . v = 72-6

u . v = 66

β•‘uβ•‘=√(u1Β²+u2Β²)

β•‘uβ•‘=√((8)Β²+(-2)Β²)

β•‘uβ•‘=√(64+4)

β•‘uβ•‘=√(68)

β•‘uβ•‘=√((4)(17))

β•‘uβ•‘=√(4)√(17)

β•‘uβ•‘=2√(17)

β•‘vβ•‘=√(v1Β²+v2Β²)

β•‘vβ•‘=√((9)Β²+(3)Β²)

β•‘vβ•‘=√(81+9)

β•‘vβ•‘=√(90)

β•‘vβ•‘=√((9)(10))

β•‘vβ•‘=√(9)√(10)

β•‘vβ•‘=3√(10)

Replacing the known values in the formula of dot product (1):

u . v =β•‘uβ•‘β•‘vβ•‘cos Ξ±

66 = 2√(17) 3√(10) cos α

Multiplying:

66 = 6√((17)(10)) cos α

66 = 6√(170) cos α

Solving first for cos α: Dividing both sides of the equation by 6√(170):

Simplifying: Dividing the numerator and denominator on the left side of the equation by 6:

(66/6)/(6√170/6)=cosΞ±β†’11/√170=cosΞ±β†’cosΞ±=11/√170

cosΞ±=11/13.03840481β†’cosΞ±=0.84366149

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