Question

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

Answer 1

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
= ...

Answer 2

MATHS

Answer 3

A village r is 10km from point p.another village a is 6km from p on a bearing 162 degree.calculate

Answer 4

34

Answer 5

using the cosine rule
100+36-120cos137degrees
136-120(-0.7313) 136+87.756=223.756.square root of 223.756=14.958=15km

Answer 6

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.

Answer 7

I don't understand pls

Answer 8

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

Answer 9

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

Answer 10

The diagram

Answer 11

The diagram