MAT Exam 2025 Paper

Concept:Use the tangent ratio in two right triangles formed at different distances from the bank to solve for the height $h$ and the river width $x$.Explanation:Let the height of the opposite point be $h$ and the width of the river be $x$.From the first position on the bank, the angle of elevation is $60^{\circ}$.Therefore, $\tan 60^{\circ} = \frac{h}{x}$.Since $\tan 60^{\circ} = \sqrt{3}$, we get $h = \sqrt{3}x$. $...(1)$After moving $30$ m away from the bank, the distance from the base becomes $x + 30$.The angle of elevation at this new position is $30^{\circ}$.Therefore, $\tan 30^{\circ} = \frac{h}{x+30}$.Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$, we get $\frac{1}{\sqrt{3}} = \frac{h}{x+30}$. $...(2)$Substitute $h = \sqrt{3}x$ from equation $(1)$ into equation $(2)$:$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}x}{x+30}$.Cross-multiply: $x + 30 = 3x$.Thus, $30 = 2x$, giving $x = 15$ m.Now substitute $x = 15$ into $h = \sqrt{3}x$:$h = 15\sqrt{3} = 15 \times 1.732 = 25.98$ m.Answer:Height of the point is $25.98$ m (approx) and the width of the river is $15$ m. Hence, option D is correct.