51. We choose +x east and +y north and measure all angles in the “standard” way (positive ones counter-

clockwise from +x, negative ones clockwise). Thus, vector 

d

1

has magnitude d

1

= 3.66 (with the unit

meter and three significant figures assumed)and direction θ

1

= 90

◦

. Also, 

d

2

has magnitude d

2

= 1.83

and direction θ

2

=

−45

◦

, and vector 

d

3

has magnitude d

3

= 0.91 and direction θ

3

=

−135

◦

. We add the

x and y components, respectively:

x :

d

1

cos θ

1

+ d

2

cos θ

2

+ d

3

cos θ

3

= 0.651 m

y :

d

1

sin θ

1

+ d

2

sin θ

2

+ d

3

sin θ

3

= 1.723 m .

(a)The magnitude of the direct displacement (the vector sum 

d

1

+ 

d

2

+ 

d

3

)is

√

0.651

2

+ 1.723

2

= 1.84 m.

(b)The angle (understood in the sense described above)is tan

−1

(1.723/0.651)= 69

◦

. That is, the first

putt must aim in the direction 69

◦

north of east.