Answer:
a. (i) the distance of R from P is 13 Km to the nearest kilometer
(ii) the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree
b. the distance of m from P is 11 Km to the nearest kilometre
Explanation:
a) A triangle PQR is formed. Q = 90°, p = 8 km; r = 10 km; distance of R from P is q is to be found.
(i) Using the cosine rule: q² = p² + r² - 2prCosQ
q² = 8² + 10² - 2 * 8 * 10 * Cos90
q² = 64 + 100 + 0
q² = 164
q = 13 Km to the nearest kilometre
Therefore, the distance of R from P is 13 Km to the nearest kilometer
(ii) the bearing of R from P
The angle at P is found using the formula Cos P = (q² + r² - p²)/2qr
Cos P = 13² + 10² - 8²/2 * 13 *10
Cos P = 0.7884
P = Cos⁻¹ 0.7884
P = 38°
Therefore, the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree
Note : 25° is alternate (Northwest) to 65°at P
b) A right-angled triangle QMP is formed
using the trigonometrical ratios; cos Θ = adjacent/hypotenuse
where the hypotenuse side = 10 km, adjacent side = distance of M from P, x
cos P = x/10
x = cos 38 * 10
x = 11 Km to the nearest kilometre
Therefore, the distance of m from P is 11 Km to the nearest kilometre
