bearings question (1 Viewer)

[= Jennifer =]

hey guys i have been trying to do this question though i can only get the answer for part b maybe amongst the highly talented students we have hear someone would be able to tell me how to get the answer

here's the question

town X is 130 km west and 210 km north of town Y. find the bearing of

a) Y from X
b) X from Y

my working out for b

N.B. theta is represented by the box
Tan  = 130
210
 = 31.75948008
= 3145
=32
therefore 360- 32 = 328 T

i just cant get part a)

ANSWERS ARE: a) 148 T b) 328 T

n.b. this question came from the insight general maths book page 176 question 17

Goodluck!!

 

[martin310015]

i don't think the answer is 148T i got something like 122T
i put it as tan@=210/130.......got something round off to 58 degree then its 180-58=122....

 

[ND]

For a) Bearing = 180-arctan(13/21)
b) Bearing = 270+arctan(21/13)

Note: arctan means inverse tan.

 

[martin310015]

arrrr shit mixed it up (checking).................no wonder i was wrong the the diagram was way off.......have to be careful next time.....

 

[= Jennifer =]

Originally posted by ND
For a) Bearing = 180-arctan(13/21)
b) Bearing = 270+arctan(21/13)

Note: arctan means inverse tan.

Originally posted by ND
For a) Bearing = 180-arctan(13/21)
b) Bearing = 270+arctan(21/13)

Note: arctan means inverse tan.

could u post ur diagram please because i think mine is wrong