(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.

Take Free Practice Test On 2025 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE

(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 °

≊ 70 °

(to the nearest degree).
(b)(i)
(ii) By the cosine rule,

| Q R |^{2} = 32^{2} + 24^{2} - 2 \times 32 \times 24 \times \cos 45

| Q R |^{2} = 1024 + 576 - 1536 \cos 45

1600 - 1086.1056

| Q R |^{2} = 513.8944

| Q R | = \sqrt{513.8944} = 22.669 k m

≊ 22.67 k m

( 2 decimal place)
(iii) By the sine rule,

\frac{32}{\sin α} = \frac{22.67}{\sin 45}

\sin α = \frac{32 \times \sin 45}{22.67}

0.9981

α = \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).

Prev Current Next

Register • Login