BEARINGS Question 12 (SSCE 1994 May/June)
(a) A village P is 10km from a lorry station, Q, on
a bearing 065º. Another village R, is 8km from
Q on a bearing 155º. Calculate
(i) the distance of R from P to the nearest
kilometer
(ii) the bearing of R from P to the nearest degree
(b) M is a village on PR such that QM is
perpendicular to PR. Find the distance of M
from P to the nearest kilometer.

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

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.