WASSCE 1995 Mathematics Question 5 Solution

WASSCE 1995 Mathematics Question 5 Problem Statement

The problem originates from WASSCE 1995, Memo No. 0160, Question 5. Points $P$, $Q$, and $R$ are situated on a flat horizontal plane. The three-figure bearing of point $Q$ measured from point $P$ is $150^\text{o}$, and the three-figure bearing of point $R$ measured from point $Q$ is $060^\text{o}$. The straight-line distance between point $P$ and point $Q$ is given as $PQ = 5\,m$, and the distance between point $Q$ and point $R$ is given as $QR = 3\,m$. The objective of the problem is to determine the bearing of point $R$ from point $P$, correct to the nearest degree.

Geometric Principles and Angle Analysis of Bearings

To establish the geometric configuration formed by points $P$, $Q$, and $R$, the properties of three-figure bearings are applied using north vertical reference lines. At point $P$, the north line points directly upwards at $000^\text{o}$, placing the direction of point $Q$ at $150^\text{o}$ clockwise from north. This implies that the line segment $PQ$ makes an angle of $180^\text{o} - 150^\text{o} = 30^\text{o}$ with the south direction line at point $P$.

By extending the north-south directional lines through point $Q$, the alternate interior angle relationship indicates that the line extending from $Q$ back to $P$ makes an angle of $30^\text{o}$ west of the north line at $Q$, corresponding to a back bearing of $330^\text{o}$. From point $Q$, point $R$ is located at a bearing of $060^\text{o}$, which is $60^\text{o}$ east of the north line at point $Q$. Adding the $30^\text{o}$ angle between line $QP$ and the north line to the $60^\text{o}$ angle between the north line and line $QR$ reveals that the interior angle $\angle PQR$ equals $30^\text{o} + 60^\text{o} = 90^\text{o}$. Therefore, triangle $PQR$ forms a right-angled triangle with the right angle located precisely at vertex $Q$

Mathematical Calculation of the Internal Angle

Since triangle $PQR$ is a right-angled triangle with $\angle PQR = 90^\text{o}$, standard right-triangle trigonometry is utilized to find the interior angle at vertex $P$, denoted as $\theta$. The side opposite to angle $\theta$ inside triangle $PQR$ is side $QR = 3\,m$, while the adjacent side to angle $\theta$ is side $PQ = 5\,m$. Applying the tangent trigonometric ratio yields $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$.

Substituting the numerical values into the formula produces $\tan(\theta) = \frac{3}{5}$, which simplifies to the decimal value $\tan(\theta) = 0.6$. To solve for angle $\theta$, the inverse tangent function is evaluated as $\theta = \tan^{-1}(0.6)$. Calculating this value gives $\theta \approx 30.96^\text{o}$, which rounds to $31^\text{o}$ when rounded to the nearest degree.

Determination of the Final Bearing of R from P

The bearing of point $R$ from point $P$ represents the total clockwise angle from the north reference line at point $P$ to the line segment $PR$. Because point $R$ lies between the north-east direction and line segment $PQ$, the line segment $PR$ subtends an interior angle of $\theta = 31^\text{o}$ inside triangle $PQR$ relative to line segment $PQ$.

The bearing of $R$ from $P$ is calculated by subtracting the interior angle $\theta$ from the bearing of point $Q$ from point $P$. Using the initial bearing of $150^\text{o}$ for point $Q$, the calculation is expressed as $180^\text{o} - 30^\text{o} - 31^\text{o}$, which equals $150^\text{o} - 31^\text{o}$. Evaluating this arithmetic expression yields $119^\text{o}$. Thus, the bearing of point $R$ from point $P$, correct to the nearest degree, is $119^\text{o}$.