JEE PYQ: Motion in a Plane - Question ID 79281a67730e (JEE Main 2020)

In problems involving angles of elevation and horizontal distances, the key concept is trigonometry in right-angled triangles. Specifically:

• The tangent of an angle in a right triangle is the ratio of the opposite side (height) to the adjacent side (horizontal distance).
• For an observer at a point on the ground, the angle $\theta$ that a vertically rising object makes with the vertical line from the observer satisfies: $\tan \theta = \frac{\text{horizontal distance from observer to vertical line}}{\text{height of object}}$
• When the observer moves horizontally, the horizontal distance changes, but the vertical motion of the object alters the height, changing the angle of elevation.

Here, we use the tangent function to relate the angles $45^\circ$ and $60^\circ$ to the heights $h_1$ and $h_1 + h_2$, and the horizontal distances $d$ and $d + 2.464d = 3.464d$.

Step 1: Define variables and initial setup

Let:

• $h_1$ = height of balloon when seen at $45^\circ$ from point B.
• $d$ = horizontal distance from point A (directly below balloon) to point B.
• When balloon rises further by $h_2$, its new height = $h_1 + h_2$.
• The girl moves further away by $2.464d$, so her new horizontal distance from A = $d + 2.464d = 3.464d$ (point C).

Step 2: Use tangent relation for first observation ($45^\circ$)

At point B, the angle with the vertical is $45^\circ$. The horizontal distance from B to A is $d$, and the height is $h_1$. Since $\tan 45^\circ = 1$, we have: $\tan 45^\circ = \frac{d}{h_1} = 1 \implies h_1 = d$

Step 3: Use tangent relation for second observation ($60^\circ$)

At point C, the angle with the vertical is $60^\circ$. The horizontal distance from C to A is $3.464d$, and the height is $h_1 + h_2 = d + h_2$. We know: $\tan 60^\circ = \sqrt{3} \approx 1.732$ So: $\tan 60^\circ = \frac{3.464d}{d + h_2} = \sqrt{3}$

Step 4: Solve for $h_2$

From the above: $\frac{3.464d}{d + h_2} = \sqrt{3} \implies 3.464d = \sqrt{3}(d + h_2) \implies d + h_2 = \frac{3.464d}{\sqrt{3}}$ We are given $\tan 30^\circ = 0.5774$, which implies $\sqrt{3} = \frac{1}{0.5774} \approx 1.732$. But more precisely, $\sqrt{3} = 1.73205$, and $3.464 = 2 \times 1.732 = 2\sqrt{3}$. So: $3.464 = 2\sqrt{3}$ Thus: $d + h_2 = \frac{2\sqrt{3} d}{\sqrt{3}} = 2d \implies h_2 = 2d - d = d$

Step 5: Conclusion

Therefore, $h_2 = d$, which corresponds to option B.

Common Mistakes:

• Students often confuse the angle with the vertical vs. the angle with the horizontal. Here, the angle is with respect to the vertical, so $\tan \theta = \frac{\text{horizontal distance}}{\text{height}}$, not $\frac{\text{height}}{\text{horizontal distance}}$.
• Misinterpreting the movement of the observer: the girl moves further by $2.464d$, so the total distance from A becomes $d + 2.464d = 3.464d$, not $2.464d$.
• Using approximate values without recognizing that $3.464 = 2\sqrt{3}$ leads to exact simplification. Students who plug in decimal approximations may lose precision.

Always draw a clear diagram labeling all distances and angles. Use exact trigonometric values (like $\sqrt{3}$) instead of decimal approximations when possible to avoid rounding errors. Double-check whether the angle is with respect to the vertical or horizontal—this changes the tangent relation.