You're on a roll — The first question (1) cannot be solved as the diagram or values for triangle LMN are not visible in the image. Question 2(a): Given triangle PQR with $PQ = 16.1\, \text{m}$, $\angle QPR = 42^\circ$, and $\angle PQR = 53^\circ$. Step 1: Find the third angle, $\angle PRQ$. The sum of angles in a triangle is $180^\circ$. $$ \angle PRQ = 180^\circ - \angle QPR - \angle PQR $$ $$ \angle PRQ = 180^\circ - 42^\circ - 53^\circ $$ $$ \angle PRQ = 180^\circ - 95^\circ $$ $$ \angle PRQ = 85^\circ $$ (i) Calculate the length of QR We use the Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B}$. $$ \frac{QR}{\sin(\angle QPR)} = \frac{PQ}{\sin(\angle PRQ)} $$ Substitute the known values: $$ \frac{QR}{\sin(42^\circ)} = \frac{16.1\, \text{m}}{\sin(85^\circ)} $$ Solve for QR: $$ QR = \frac{16.1 \times \sin(42^\circ)}{\sin(85^\circ)} $$ $$ QR = \frac{16.1 \times 0.6691}{0.9962} $$ $$ QR = \frac{10.77551}{0.9962} $$ $$ QR \approx 10.816\, \text{m} $$ Rounding to three significant figures: $$ \boxed{10.8\, \text{m}} $$ (ii) Calculate the Area of triangle PQR We use the formula for the area of a triangle given one side and all angles: Area $= \frac{1}{2} \frac{c^2 \sin A \sin B}{\sin C}$. Here, $c = PQ = 16.1\, \text{m}$, $A = \angle QPR = 42^\circ$, $B = \angle PQR = 53^\circ$, and $C = \angle PRQ = 85^\circ$. $$ \text{Area} = \frac{1}{2} \frac{(PQ)^2 \sin(\angle QPR) \sin(\angle PQR)}{\sin(\angle PRQ)} $$ $$ \text{Area} = \frac{1}{2} \frac{(16.1)^2 \sin(42^\circ) \sin(53^\circ)}{\sin(85^\circ)} $$ $$ \text{Area} = \frac{1}{2} \frac{259.21 \times 0.6691 \times 0.7986}{0.9962} $$ $$ \text{Area} = \frac{1}{2} \frac{138.471}{0.9962} $$ $$ \text{Area} = \frac{1}{2} \times 139.000 $$ $$ \text{Area} \approx 69.500\, \text{m}^2 $$ Rounding to three significant figures: $$ \boxed{69.5\, \text{m}^2} $$ (iii) Calculate the Shortest distance from R to PQ The shortest distance from R to PQ is the altitude (height) from