MAT Exam 2023 Paper

Concept:Use $\tan$ ratios for the two angles of elevation and apply distance = speed × time.Explanation:Let the initial distance of the person from the building be $x$ m.Distance travelled in 2 hours:$$d = 25(\sqrt{3}-1) \times 2 = 50(\sqrt{3}-1)\ \text{m}$$From the first observation,$$\tan 30^\circ = \frac{h}{x}$$$$\frac{1}{\sqrt{3}} = \frac{h}{x} \Rightarrow h = \frac{x}{\sqrt{3}}$$After moving closer, the remaining distance is $x - 50(\sqrt{3}-1)$.From the second observation,$$\tan 45^\circ = \frac{h}{x - 50(\sqrt{3}-1)}$$$$1 = \frac{h}{x - 50(\sqrt{3}-1)} \Rightarrow h = x - 50(\sqrt{3}-1)$$Equating both expressions for $h$:$$\frac{x}{\sqrt{3}} = x - 50(\sqrt{3}-1)$$$$x - \frac{x}{\sqrt{3}} = 50(\sqrt{3}-1)$$$$\frac{x(\sqrt{3}-1)}{\sqrt{3}} = 50(\sqrt{3}-1)$$Cancelling $(\sqrt{3}-1)$:$$\frac{x}{\sqrt{3}} = 50 \Rightarrow x = 50\sqrt{3}$$Substitute $x$ into $h = \frac{x}{\sqrt{3}}$:$$h = \frac{50\sqrt{3}}{\sqrt{3}} = 50\ \text{m}$$Answer:The height of the building is $50$ m, i.e., option C.