Advanced Heights and Distances: Lake Reflections and Speed for CBSE Class 10

In CBSE Class 10 Mathematics board examinations, Section D contains the highest-weightage, challenging 5-mark questions. Within trigonometry, these problems go beyond static observations: they involve dynamic moving objects (airplanes and cars changing elevation angles over time) or optical reflection physics where the virtual image of an object lies as far below a water surface as the object lies above it.

Mastering these two advanced archetypes—the Cloud and Lake Reflection problem and the Moving Aeroplane / Car Speed problem—prepares students to solve the most difficult questions that can appear on the board exam with total confidence.

What You Will Learn

• Optical reflection geometry: Real object distance above water = Virtual image distance below water
• Complete, rigorous proof of the Cloud and Lake Reflection Theorem: H = rac{h( an eta + an lpha)}{ an eta - an lpha}
• Step-by-step method for Moving Aeroplane Speed Problems ( ext{Speed} = rac{ ext{Distance}}{ ext{Time}})
• The Car Approaching a Tower Problem (calculating time to reach base using uniform speed)
• Board exam presentation templates and algebraic manipulation strategies

1. The Cloud and Lake Reflection Problem (CBSE Most Celebrated 5-Mark Question)

Problem Statement:

The angle of elevation of a cloud from a point $h$ metres above a lake is lpha, and the angle of depression of its reflection in the lake is eta. Prove that the height of the cloud above the lake surface is: \mathbf{H = rac{h( an eta + an lpha)}{ an eta - an lpha}}

                                  C (Cloud)
                                  |
                                  | H - h
             Observation Point A -+------------------ M (Horizontal Reference)
                               |  |\ α
                             h |  |                                |  |  \ β
             Lake Surface -----+--+---+-------------- L
                                  | H
                                  |
                                  |
                                  C' (Virtual Reflection of Cloud)
                                  (Distance below lake = H)

Step-by-Step Proof:

1. Understand the Physics of Reflection:

• Let the surface of the lake be represented by the horizontal line $L$.
• Let the true height of the cloud $C$ above the lake surface be $H ext{ metres}$.
• By the laws of optical reflection in plane mirrors, the virtual image of the cloud in the lake ($C'$) lies at the exact same distance below the lake surface as the cloud is above it: $ext{Depth of reflection } C' \text{ below lake surface} = \mathbf{H ext{ metres}}$

2. Establish the Elevated Reference Frame:

• The observer is at point $A$, at a height of $h ext{ metres}$ above the lake surface.
• Draw a horizontal line $AM$ through $A$.
• Distance from horizontal line $AM$ up to the cloud $C$: $CM = H - h$
• Distance from horizontal line $AM$ down to the reflection $C'$: $C'M = ( ext{Depth to lake surface}) + ( ext{Height of } A ext{ above lake}) = \mathbf{H + h}$
• Let the horizontal distance $AM$ be $x ext{ metres}$.

3. In Right Triangle $\Delta AMC$ (Angle of Elevation lpha):

an lpha = rac{CM}{AM} = rac{H - h}{x} x = rac{H - h}{ an lpha} \quad ext{--- (1)}

4. In Right Triangle $\Delta AMC'$ (Angle of Depression eta):

an eta = rac{C'M}{AM} = rac{H + h}{x} x = rac{H + h}{ an eta} \quad ext{--- (2)}

5. Equate Equations (1) and (2) to Eliminate $x$:

rac{H - h}{ an lpha} = rac{H + h}{ an eta} Cross-multiply: (H - h) an eta = (H + h) an lpha H an eta - h an eta = H an lpha + h an lpha

6. Group Terms Containing $H$ on LHS and $h$ on RHS:

H an eta - H an lpha = h an eta + h an lpha Factor out $H$ on LHS and $h$ on RHS: H( an eta - an lpha) = h( an eta + an lpha)

Dividing both sides by ( an eta - an lpha): \mathbf{H = rac{h( an eta + an lpha)}{ an eta - an lpha}} Hence, proved.

2. Dynamic Speed Problems: The Moving Aeroplane

In dynamic problems, an object moves horizontally at a constant altitude, causing its angle of elevation from a fixed ground station to decrease over time.

Solved Example: Speed of an Aeroplane (NCERT Classic)

Problem: The angle of elevation of an aeroplane from a point on the ground is $60^\circ$. After a flight of $30 ext{ seconds}$, the angle of elevation changes to $30^\circ$. If the aeroplane is flying at a constant height of $3600\sqrt{3} ext{ m}$, find the speed of the aeroplane in $ext{km/h}$.

             P (Plane at t = 0) ------------- Q (Plane at t = 30s)
                     |       Distance d = PQ       |
         3600√3 m    |                             | 3600√3 m
                     |                             |
     Ground Point O -+-----------------------------+
                     A                             B

Solution:

1. Geometric Setup:
   • Let $O$ be the fixed observation point on the ground.
   • Initial position of plane: $P$, at height $PA = 3600\sqrt{3} ext{ m}$. Angle of elevation ngle POA = 60^\circ.
   • Position of plane after $30 ext{ seconds}$: $Q$, at height $QB = 3600\sqrt{3} ext{ m}$. Angle of elevation ngle QOB = 30^\circ.
   • Horizontal distance flown in 30 seconds: $d = PQ = AB = OB - OA$.
2. From Right Triangle $\Delta OAP$: an 60^\circ = rac{PA}{OA} \implies \sqrt{3} = rac{3600\sqrt{3}}{OA} OA = rac{3600\sqrt{3}}{\sqrt{3}} = \mathbf{3600 ext{ m}}
3. From Right Triangle $\Delta OBQ$: an 30^\circ = rac{QB}{OB} \implies rac{1}{\sqrt{3}} = rac{3600\sqrt{3}}{OB} $OB = 3600\sqrt{3} imes \sqrt{3} = 3600 imes 3 = \mathbf{10800 ext{ m}}$
4. Calculate Distance Flown ($d$): $d = OB - OA = 10800 - 3600 = \mathbf{7200 ext{ m}}$
5. Calculate Speed of the Plane: ext{Speed} = rac{ ext{Distance}}{ ext{Time}} = rac{7200 ext{ m}}{30 ext{ s}} = \mathbf{240 ext{ m/s}}
6. Convert Speed to $ext{km/h}$: ext{Speed in km/h} = 240 imes rac{18}{5} = 48 imes 18 = \mathbf{864 ext{ km/h}}
7. Therefore, <u>the speed of the aeroplane is $864 ext{ km/h}$</u>.

3. Dynamic Speed Problems: Car Approaching a Tower (NCERT Classic)

Problem: A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30^\circ$, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be $60^\circ$. Find the time taken by the car to reach the foot of the tower from this point.

Solution:

1. Let $AB = h$ be the tower, $C$ be the position of the car at $t = 0$ (ngle ACB = 30^\circ), and $D$ be the position of the car 6 seconds later (ngle ADB = 60^\circ).
2. Let uniform speed of car be $v ext{ m/s}$. Distance $CD = ext{Speed} imes ext{Time} = 6v ext{ metres}$. Let time taken from $D$ to foot $B$ be $t ext{ seconds} \implies DB = vt ext{ metres}$. Total distance $CB = 6v + vt = v(6 + t)$.
3. In Right Triangle $\Delta ABD$: an 60^\circ = rac{h}{DB} \implies \sqrt{3} = rac{h}{vt} \implies h = vt\sqrt{3} \quad ext{--- (1)}
4. In Right Triangle $\Delta ABC$: an 30^\circ = rac{h}{CB} \implies rac{1}{\sqrt{3}} = rac{h}{v(6 + t)} \implies h = rac{v(6 + t)}{\sqrt{3}} \quad ext{--- (2)}
5. Equate Expressions for $h$: vt\sqrt{3} = rac{v(6 + t)}{\sqrt{3}} Divide both sides by $v$ ($v e 0$): $t\sqrt{3} imes \sqrt{3} = 6 + t$ $3t = 6 + t \implies 2t = 6 \implies t = \mathbf{3 ext{ seconds}}$
6. Therefore, <u>the time taken by the car to reach the foot of the tower from that point is $3 ext{ seconds}$</u>.

4. Summary and Examination Tips

Exam Tip: In the Cloud and Lake reflection problem, students often mistakenly write the depth of the reflection from the observer as $H$. Always state explicitly: "Depth of reflection from lake surface is $H$, hence distance from the horizontal line of observation is $H + h$".

Common Mistake: In speed problems, giving the final answer in $ext{m/s}$ when the question asked for $ext{km/h}$. Always check the requested units in the final sentence!