You can just draw the triangle, and use the cosine law.
let the distance from P to R be d
d^2 = 20^2 + 15^2 - 2(20)(15)cos135°
= 1049.264...
d = appr 32.39 km
The angle between North and PR:
sinP/15 = sin135/32.39..
sinP = .32744..
angle P = 19.11°
state the answer using the "bearing" notation that you learned or use
Same solution using vectors
vector PR = (20cos90, 20sin90) + (15cos45, 15sin45)
= (0, 20) + (10.6066.. , 10.6066.. )
= (10.6066.. ,30.6066..)
resultant = √(10.6066..^2 + 30.6066..^2) = 32.39 , just as before
for the angle at P
tanP = 30.6066../10.6066.. = 2.8856
angle P = 70.886° <---- from the x-axis , standard trig notation
so the angle between North and PR = 90° - 70.86° = 19.11°
again, just as before.
A car leaves P and drive 20km north to Q, from Q it drive 15km on a bearing N45° E to R.find the distance and bearing of R from P.
3 answers
Vgggjnnh
I'm sorry, I don't understand what you are trying to say. Can you please rephrase or provide more context?