=13 × 2+6×2-2(13)(6) cos99
=169+36+24.4038
b×b=229.4038
b=√229.4038
=15.14608
=15km
A student walks 50m on a bearing of 25 and then 200m due east. how far is she from her starting point
2 answers
Your response to your own question has no resemblance at all
to anything in your problem. No idea what it is supposed to do or say.
I made a sketch and have triangle with sides 50 and 200 with an angle
of 115° between those sides, so, clearly I will use the cosine law.
d^2 = 50^2 + 200^2 - 2(50)(200)cos115°
= 50,952.365..
d = appr 225.7 m
Or
we could use vectors:
vector d = 50(cos65,sin65) + 200(cos0,sin0)
= (21.131,45.3154) + (200,0)
= (221.131, 45.3154)
|d| = √(221.131^2 + 45.3154^2)
= appr 225.7 , just like before
to anything in your problem. No idea what it is supposed to do or say.
I made a sketch and have triangle with sides 50 and 200 with an angle
of 115° between those sides, so, clearly I will use the cosine law.
d^2 = 50^2 + 200^2 - 2(50)(200)cos115°
= 50,952.365..
d = appr 225.7 m
Or
we could use vectors:
vector d = 50(cos65,sin65) + 200(cos0,sin0)
= (21.131,45.3154) + (200,0)
= (221.131, 45.3154)
|d| = √(221.131^2 + 45.3154^2)
= appr 225.7 , just like before