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.
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.
Vgggjnnh
I'm sorry, I don't understand what you are trying to say. Can you please rephrase or provide more context?
To find the distance and bearing of R from P, you can use the concept of vector addition and trigonometry.
First, let's break down the given information into vectors:
1. The car initially travels 20 km north from P to Q. We can represent this as a vector from P to Q: PQ = 20 km.
2. From Q, the car then travels 15 km on a bearing of N45° E to reach R. The bearing means the angle measured clockwise from the north direction. In this case, N45° E means a bearing of 45° east of north.
To find the distance and bearing of R from P, we need to add the vectors PQ and QR. Since Q is the starting point for the second vector, we need to express QR in terms of its north and east components.
Now, let's calculate the components of QR:
The north component of QR can be calculated by multiplying the distance (15 km) by the sine of the angle between QR and the north direction. Since QR is N45° E, the angle between QR and north is 45°. Therefore, the north component of QR is 15 km * sin(45°).
The east component of QR can be calculated by multiplying the distance (15 km) by the cosine of the angle between QR and the north direction. In this case, since QR is N45° E, the angle between QR and north is 45°, and the east component of QR is 15 km * cos(45°).
Next, let's calculate the north and east components of QR:
North component = 15 km * sin(45°) ≈ 10.6 km (rounded to 1 decimal place)
East component = 15 km * cos(45°) ≈ 10.6 km (rounded to 1 decimal place)
Now, we can calculate the total distance and direction from P to R:
Distance from P to R = Distance from P to Q + Distance from Q to R
Distance from P to R = PQ + QR
Distance from P to R = 20 km + √(North component of QR)^2 + (East component of QR)^2
Distance from P to R = 20 km + √(10.6 km)^2 + (10.6 km)^2
Distance from P to R ≈ 20 km + √(112.36 km^2 + 112.36 km^2) ≈ 20 km + √(224.72 km^2) ≈ 20 km + 14.99 km ≈ 34.99 km (rounded to 2 decimal places)
Therefore, the distance from P to R is approximately 34.99 km.
To find the bearing of R from P, we can use the inverse tangent (arctan) function to determine the angle between the north direction and the line connecting P and R.
Bearing of R from P = arctan(East component of QR / North component of QR)
Bearing of R from P = arctan(10.6 km / 10.6 km) = arctan(1) = 45°
Therefore, the bearing of R from P is N45° E.