(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.
(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.
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Correct Answer: Option n
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).
(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).