(a) Given that (5 cos (x + 8.5)° – 1 = 0, 0° leq x leq 90°), calculate, correct to the nearest degree, the value of x.
(b) The bearing of Q from P is 0150° and the bearing of P from R is 015°. If Q and R are 24km and 32km respectively from P : (i) represent this information in a diagram;
(ii) calculate the distance between Q and R, correct to two decimal places ; (iii) find the bearing of R from Q, correct to the nearest degree.
Explanation
(a) (5 cos (x + 8.5)° – 1 = 0)
(5 cos (x + 8.5)° = 1)
(cos (x + 8.5)° = frac{1}{5} = 0.2)
((x + 8.5)° = cos^{-1}(0.2) = 78.463°)
(x = 78.463° – 8.5 = 69.963°)
(approxeq 70°) (to the nearest degree).
(b)(i)
(ii) By the cosine rule,
(|QR|^{2} = 32^{2} + 24^{2} – 2 times 32 times 24 times cos 45)
(|QR|^{2} = 1024 + 576 – 1536 cos 45)
= (1600 – 1086.1056)
(|QR|^{2} = 513.8944)
(|QR| = sqrt{513.8944} = 22.669 km)
(approxeq 22.67 km) ( 2 decimal place)
(iii) By the sine rule,
(frac{32}{sin alpha} = frac{22.67}{sin 45})
(sin alpha = frac{32 times sin 45}{22.67})
= (0.9981)
(alpha = sin^{-1} (0.9981) = 86.4787°)
The diagram below shows all the angles at Q;
The bearing of R from Q is given by the reflex angle NQR. Thus
reflex < NQR = 360° – (86.47° + 30°) = 360° – 116.47°
= 243.53°
Hence, the bearing of R from Q = 244° (to the nearest degree).