MAT Exam 2020 Paper

Concept:Use the tangent ratio $\tan \theta = \frac{\text{height}}{\text{distance from tower}}$ for two positions and subtract to eliminate the unknown closer distance.Explanation:Let the height of the tower be $h$ metres.Let the closer point be $Q$, with $QR$ as its distance from the tower base.The first point is $192$ m farther away, so its distance from the tower base is $192 + QR$.At the first point:$$\tan \alpha = \frac{h}{192 + QR} = \frac{5}{12}$$Taking reciprocals:$$\frac{192 + QR}{h} = \frac{12}{5}$$At the closer point:$$\tan \beta = \frac{h}{QR} = \frac{3}{4}$$So:$$\frac{QR}{h} = \frac{4}{3}$$Subtract the second equation from the first:$$\frac{192}{h} = \frac{12}{5} - \frac{4}{3}$$$$\frac{192}{h} = \frac{36 - 20}{15} = \frac{16}{15}$$Therefore:$$h = \frac{192 \times 15}{16} = 180 \text{ m}$$Answer:The height of the tower is $180$ m.Correct option: C.