Q.An aircraft is flying at a height of $3400\ \text{m}$ above the ground. If the angle subtended at a ground observation point by the aircraft positions $10.0\ \text{s}$ apart is $30^\circ$, what is the speed of the aircraft?

The key idea is to relate the aircraft's linear motion to the observed angular displacement using trigonometry. The rate of change of this angular displacement is the angular velocity of the line of sight.

1. Assume the observer is positioned such that the aircraft's path segment is symmetric with respect to the observer's vertical line. This forms an isosceles triangle with the observer at the apex and the aircraft's two positions as base vertices.
2. Draw a perpendicular from the observer to the aircraft's path. This line has length $h = 3400\ \text{m}$ and bisects the angle subtended ($30^\circ$) and the horizontal distance traveled ($d$).
3. In the resulting right-angled triangle, half the subtended angle is $30^\circ/2 = 15^\circ$. We can write $\tan(15^\circ) = \frac{d/2}{h}$.
4. Solve for $d$: $d = 2h \tan(15^\circ)$. We know $\tan(15^\circ) = 2 - \sqrt{3}$. …

The problem asks for the linear speed of an aircraft, given its height, the time interval, and the angular displacement it subtends at a ground observation point. The core idea is to translate this angular information into a linear distance using the geometry of the situation, specifically trigonometry.

Imagine the aircraft at two different positions, $P_1$ and $P_2$, $10.0\ \text{s}$ apart. An observer at $O$ on the ground sees these two positions, and the angle formed by the lines of sight $OP_1$ and $OP_2$ is $30^\circ$. Since the aircraft flies at a constant height, we can construct a right-angled triangle that connects the aircraft's height, the linear distance it travels, and the observed angle. Once we find the linear distance traveled, calculating the speed is straightforward.

1. Visualize the Geometry:

   Let the observer be at point $O$ on the ground. The aircraft flies horizontally at a constant height $h = 3400\ \text{m}$. Let $P_1$ and $P_2$ be the two positions of the aircraft $10.0\ \text{s}$ apart. The angle subtended at the observation point $O$ by these two positions is $\angle P_1 O P_2 = 30^\circ$.

   For problems of this type, it is standard to assume the observer is positioned such that the aircraft's path is symmetric with respect to the observer's line of sight. This means the observer $O$ is directly below the midpoint of the segment $P_1 P_2$.

   Let $M$ be the midpoint of the segment $P_1 P_2$. Then the line segment $OM$ represents the perpendicular distance from the observer to the aircraft's path, which is equal to the height $h$.

   This setup forms an isosceles triangle $\triangle P_1 O P_2$, where $OP_1 = OP_2$.

             P1 -------- M -------- P2  (Aircraft path at height h)
             |          |          |
             |          | h        |
             |          |          |
             -------------------------- (Ground level)
                        O (Observer)
   

2. Identify Knowns and Unknowns:

   • Height of aircraft, $h = 3400\ \text{m}$.
   • Time interval, $\Delta t = 10.0\ \text{s}$.
   • Angle subtended, $\theta = \angle P_1 O P_2 = 30^\circ$.
   • We need to find the speed of the aircraft, $v$.

3. Relate Linear Distance to Angular Displacement using Trigonometry:

   In the isosceles triangle $\triangle P_1 O P_2$, the line segment $OM$ is the altitude from $O$ to $P_1 P_2$. It bisects the angle $\angle P_1 O P_2$ and also bisects the segment $P_1 P_2$.

   Consider the right-angled triangle $\triangle O M P_2$.

   The angle $\angle M O P_2 = \frac{\theta}{2} = \frac{30^\circ}{2} = 15^\circ$.

   The side $OM$ is the height $h$.

   The side $MP_2$ is half the linear distance traveled by the aircraft. Let $d = P_1 P_2$, so $MP_2 = d/2$.

   Using the tangent function in $\triangle O M P_2$:

4. Calculate the Linear Distance ($d$): From the trigonometric relation, we can express $d$:

To find the exact value of $\tan(15^\circ)$, we use the angle subtraction formula for tangent:
$$ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} $$ …

Concept: Law of Cosines on the Whole Triangle, with $\cos15^\circ$ from the Half-Angle Formula

Method: One Law-of-Cosines Step (No Bisecting into Two Right Triangles), $\cos15^\circ$ via the Half-Angle Identity (Not Angle Subtraction)

The existing solutions bisect the isosceles triangle into two right triangles and use $\tan15^\circ$ (found via the angle-subtraction formula $\tan(45^\circ-30^\circ)$) to get half the distance. This method instead applies the Law of Cosines to the whole triangle $OP_1P_2$ directly — no bisection needed — and gets $\cos15^\circ$ from the half-angle identity instead of angle subtraction.

Step 1 — Set up the isosceles triangle

The observer $O$ is directly below the aircraft's flight line, at perpendicular distance $h=3400\ \text{m}$ from it. The two aircraft positions $P_1,P_2$ (seen $30^\circ$ apart) are equidistant from $O$: $OP_1=OP_2$. In the right triangle formed by dropping a perpendicular from $O$ to the midpoint of $P_1P_2$, the half-angle is $15^\circ$ and:

Step 2 — Get $\cos15^\circ$ from the half-angle formula (not angle subtraction)

Step 3 — Side length $OP_1=OP_2$

Step 4 — Law of Cosines on the whole triangle $OP_1P_2$ (single step, no bisection)

The distance $d=P_1P_2$ is the side opposite the $30^\circ$ angle at $O$, so directly:

(using $OP_1=OP_2$ to simplify). Substituting:

$d^2 = 2(3520.6)^2(1-0.8660) = 2(12{,}394{,}624)(0.1340) \approx 3{,}321{,}770$ …