the angle between the two directions is
162-25 or 137
Use the cosine law, let the distance between them be d
d^2 = 10^2 + 6^2 - 2(10)(6)cos137
= ...
A village R is 10km from a point P on a bearing 025 from P. Another village A is 6km from P on a bearing 162. Calculate the distance of R from A
11 answers
MATHS
A village r is 10km from point p.another village a is 6km from p on a bearing 162 degree.calculate
34
using the cosine rule
100+36-120cos137degrees
136-120(-0.7313) 136+87.756=223.756.square root of 223.756=14.958=15km
100+36-120cos137degrees
136-120(-0.7313) 136+87.756=223.756.square root of 223.756=14.958=15km
A town P is 10km from a lorry station Q on a bearing of 065 degree. Another town R is 8km from Q on a bearing of 155 degrees. Determine the distance of R from P to the nearest km and the bearing of R from P.
I don't understand pls
A village R is 10km from a point P on a bearing 025 from P. Another village A is 6km from P on a bearing 162. Calculate
a. The distance of R from A.
b. The bearing of R from A
Solution
Representing the bearing and distance on a triangle gives:
<RPA = 162 – 25 = 1370
[a] We will use Cosine Rule to determine the distance of R from A (p);
From Cosine Rule;
b2 = a2 + c2 – 2ac Cos B
p2 = a2 + r2 – 2ar Cos P
where
a= 10 km Cos P = Cos 137
r = 6km
p = ?
p2 = a2 + r2 – 2ar Cos P
p2 = 102 + 62 - 2 (10) (6) Cos 1370
p2 = 100 + 36 – 120 (- 0.7314)
p2 = 136 + 87.76
p2 = 223.76
p = 14.95 = 15km (approximated to the nearest km)
[b] To find the bearing of the R from A,
we will apply Sine Rule;
From Sine Rule;
15 * Sin Ø = 10 * Sin 137 = 6.82
Sin Ø = 6.82/15 = 0.4547
Ø = ArcSin 0.4547 = 27.0o
Therefore, the bearing of R from A will be
<PRA = 180 – 137 – 27 = 160
The alternating angle to 25o at R is 65o
Then, the bearing of R from A = 270 -65-16 = 1890
a. The distance of R from A.
b. The bearing of R from A
Solution
Representing the bearing and distance on a triangle gives:
<RPA = 162 – 25 = 1370
[a] We will use Cosine Rule to determine the distance of R from A (p);
From Cosine Rule;
b2 = a2 + c2 – 2ac Cos B
p2 = a2 + r2 – 2ar Cos P
where
a= 10 km Cos P = Cos 137
r = 6km
p = ?
p2 = a2 + r2 – 2ar Cos P
p2 = 102 + 62 - 2 (10) (6) Cos 1370
p2 = 100 + 36 – 120 (- 0.7314)
p2 = 136 + 87.76
p2 = 223.76
p = 14.95 = 15km (approximated to the nearest km)
[b] To find the bearing of the R from A,
we will apply Sine Rule;
From Sine Rule;
15 * Sin Ø = 10 * Sin 137 = 6.82
Sin Ø = 6.82/15 = 0.4547
Ø = ArcSin 0.4547 = 27.0o
Therefore, the bearing of R from A will be
<PRA = 180 – 137 – 27 = 160
The alternating angle to 25o at R is 65o
Then, the bearing of R from A = 270 -65-16 = 1890
A village R is 10km from a point P on a bearing 025 from P. Another village A is 6km from P on a bearing 162. Calculate
a. The distance of R from A.
b. The bearing of R from A
Solution
Representing the bearing and distance on a triangle gives:
<RPA = 162 – 25 = 1370
[a] We will use Cosine Rule to determine the distance of R from A (p);
From Cosine Rule;
b2 = a2 + c2 – 2ac Cos B
p2 = a2 + r2 – 2ar Cos P
where
a= 10 km Cos P = Cos 137
r = 6km
p = ?
p2 = a2 + r2 – 2ar Cos P
p2 = 102 + 62 - 2 (10) (6) Cos 1370
p2 = 100 + 36 – 120 (- 0.7314)
p2 = 136 + 87.76
p2 = 223.76
p = 14.95 = 15km (approximated to the nearest km)
[b] To find the bearing of the R from A,
we will apply Sine Rule;
From Sine Rule;
15 * Sin Ø = 10 * Sin 137 = 6.82
Sin Ø = 6.82/15 = 0.4547
Ø = ArcSin 0.4547 = 27.0degrees
Therefore, the bearing of R from A will be
<PRA = 180 – 137 – 27 = 16 degrees
The alternating angle to 25 degrees at R is 65 degrees
Then, the bearing of R from A = 270 -65-16 = 189 degrees
a. The distance of R from A.
b. The bearing of R from A
Solution
Representing the bearing and distance on a triangle gives:
<RPA = 162 – 25 = 1370
[a] We will use Cosine Rule to determine the distance of R from A (p);
From Cosine Rule;
b2 = a2 + c2 – 2ac Cos B
p2 = a2 + r2 – 2ar Cos P
where
a= 10 km Cos P = Cos 137
r = 6km
p = ?
p2 = a2 + r2 – 2ar Cos P
p2 = 102 + 62 - 2 (10) (6) Cos 1370
p2 = 100 + 36 – 120 (- 0.7314)
p2 = 136 + 87.76
p2 = 223.76
p = 14.95 = 15km (approximated to the nearest km)
[b] To find the bearing of the R from A,
we will apply Sine Rule;
From Sine Rule;
15 * Sin Ø = 10 * Sin 137 = 6.82
Sin Ø = 6.82/15 = 0.4547
Ø = ArcSin 0.4547 = 27.0degrees
Therefore, the bearing of R from A will be
<PRA = 180 – 137 – 27 = 16 degrees
The alternating angle to 25 degrees at R is 65 degrees
Then, the bearing of R from A = 270 -65-16 = 189 degrees
The diagram
The diagram
5 answers