(a) A balloon is observed simultaneously from three points A, B and C on a straight road directly be

Q: Wait, why do we reject the negative root when finding $\sin \alpha$?

In this geometric context, $\alpha$ represents an angle of elevation, which must be positive and acute (since the sum of angles in a triangle is $180^\circ$). Thus, $\sin \alpha$ must be positive.

Q: Why can't we just use the Cosine Rule?

You absolutely can! However, the Sine Rule is often more elegant when you have multiple triangles sharing a common side, as it avoids the complexity of squaring terms early on.

Q: How do I remember the circle theorem for the second part?

Think of the center of the circle as the 'source' of the angle. The angle at the center is always twice the angle at the circumference because the arc length is proportional to the central angle.

Question

(a) A balloon is observed simultaneously from three points $A, B$ and $C$ on a straight road directly beneath it. The angular elevation at $B$ is twice that at $A$ and the angular elevation at $C$ is thrice that at $A$. If the distance between $A$ and $B$ is $a$ and the distance between $B$ and $C$ is $b$, find the height of the balloon in terms of $a$ and $b$. (b) Find the area of the smaller part of a disc of radius $10$ cm, cut off by a chord $AB$ which subtends an angle of $22 \frac{1}{2}^\circ$ at the circumference.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We have a balloon at point $P$ and its projection on the road at $Q$. The points $A$, $B$, and $C$ lie on the road such that the angles of elevation are $\angle PAQ = \alpha$, $\angle PBQ = 2\alpha$, and $\angle PCQ = 3\alpha$.

Our objective is to determine the height $h = PQ$ in terms of the distances between the observation points. Let $AB = a$ and $BC = b$.

### The 'Aha!' Moment: The Exterior Angle Theorem

Consider $	riangle ABP$. The angle $\angle PBQ$ is an exterior angle to this triangle. According to the Exterior Angle Theorem, the exterior angle is equal to the sum of the two opposite interior angles.

Thus, $\angle PBQ = \angle PAQ + \angle APB$. Substituting our known values, we get $2\alpha = \alpha + \angle APB$, which implies $\angle APB = \alpha$.

Since $\angle PAB = \alpha$ and $\angle APB = \alpha$, $	riangle ABP$ is isosceles. This confirms that $PB = AB = a$.

Now, apply the same logic to $	riangle PBC$. The angle $\angle PCQ$ is an exterior angle to $	riangle PBC$, so $\angle PCQ = \angle PBQ + \angle BPC$.

Substituting the values, $3\alpha = 2\alpha + \angle BPC$, which gives us $\angle BPC = \alpha$. We now have a common angle $\alpha$ in both triangles, which serves as our mathematical bridge.

### The Power of the Sine Rule

In $	riangle PBC$, we know $\angle PCB = 180^{\circ} - 3\alpha$. Applying the Sine Rule:

$$\frac{PB}{\sin(180^{\circ} - 3\alpha)} = \frac{BC}{\sin \alpha}$$

Since $\sin(180^{\circ} - 	heta) = \sin 	heta$, this simplifies to:

$$\frac{a}{\sin 3\alpha} = \frac{b}{\sin \alpha}$$

Invoke the triple-angle identity $\sin 3\alpha = 3\sin \alpha - 4\sin^3 \alpha$. Substituting this into our equation:

$$\frac{a}{3\sin \alpha - 4\sin^3 \alpha} = \frac{b}{\sin \alpha}$$

Canceling $\sin \alpha$ (given $\alpha 
eq 0$), we arrive at:

$$\frac{a}{3 - 4\sin^2 \alpha} = b \Rightarrow \sin^2 \alpha = \frac{3b - a}{4b}$$

### Final Calculation for Height

To find the height $h$, we use $h = PB \sin 2\alpha = a(2 \sin \alpha \cos \alpha)$. We first determine $\cos^2 \alpha$:

$$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{3b - a}{4b} = \frac{b + a}{4b}$$

Substituting these into the expression for $h$:

$$h = 2a \sqrt{\frac{3b - a}{4b}} \sqrt{\frac{b + a}{4b}} = \frac{a}{2b} \sqrt{(a + b)(3b - a)}$$

### The Disc and the Chord

We have a circular disc where a chord $AB$ subtends $22 \frac{1}{2}^{\circ}$ at the circumference. A fundamental circle theorem states that the angle subtended by an arc at the center is twice the angle at the circumference.

Thus, the central angle $	heta = 2 	imes 22 \frac{1}{2}^{\circ} = 45^{\circ}$, or $\frac{\pi}{4}$ radians.

The area of the smaller segment is the area of the sector minus the area of the triangle:

$$	ext{Area} = \frac{1}{2} r^2 (	heta - \sin 	heta)$$

With $r = 10$ and $	heta = \frac{\pi}{4}$, we calculate:

$$	ext{Area} = 50 \left( \frac{\pi}{4} - \sin 45^{\circ} ight) \approx 50 (0.7854 - 0.7071) \approx \mathbf{3.91 	ext{ cm}^2}$$

Visualized Solution

Visualizing the Balloon and the Road

Analyzing Triangle $A B P$

Analyzing Triangle $P B C$

Applying Sine Rule in $△ P B C$

Solving for $sin^{2} α$

Finding $cos^{2} α$

Calculating the Height $h$

Transition to Part (b): The Disc Problem

Finding the Central Angle

Formula for Area of Segment

Final Calculation for Area

Summary and Key Takeaways

The Sigma Insight: Heights and Distances

Analyzing the Setup

The 'Aha!' Moment

The Power of the Sine Rule

Final Calculation for Height

The Disc and the Chord

Similar Questions

A spherical gas balloon of radius 16 meter subtends an angle $6 0^{\circ}$ at the eye of the observer $A$ while the angle of elevation of its center from the eye of $A$ is $7 5^{\circ}$ . Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is:

A tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $A B (= a)$ subtends an angle of $6 0^{\circ}$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is $3 0^{\circ}$ . The height of the tower is

A pole stands vertically inside a triangular park $A B C$ . Let the angle of elevation of the top of the pole from each corner of the park be $\frac{π}{3}$ . If the radius of the circumcircle of $Δ A B C$ is 2, then the height of the pole is equal to: