JEE Advanced 1985

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Animated Solution for Mathematics - Trigonometry: A ladder rests against a wall at an angle $α$ to the horizontal. Its foot is pulled away from the wall through a distance $a$ , so that it slides a distance $b$ down the wall making an angle $β$ with the horizontal. Show that $a = b tan \frac{1}{2} (α + β)$ .

Q: Wait, does the length of the ladder $\ell$ actually matter? If it cancels out, does that mean any ladder would behave the same way?

That is a brilliant observation! Yes, the length $\ell$ cancels out because the relationship between the shifts $a$ and $b$ is purely geometric—it depends only on the angular change. It implies that the 'geometry of the slide' is scale-invariant. Whether your ladder is 2 meters or 20 meters, the ratio of the shifts depends solely on the angles.

Q: Can we solve this using coordinate geometry? Let the wall be the y-axis and the floor the x-axis?

Absolutely! You can set the coordinates of the foot as $(x, 0)$ and the top as $(0, y)$. Then $x^2 + y^2 = \ell^2$. You can differentiate or use the change in coordinates. However, the trigonometric approach is often faster and more elegant for this specific problem. In JEE, choosing the path of least resistance is a skill in itself!

Progressive Hints

Hint 1

Visualize the ladder as the hypotenuse of two distinct right-angled triangles. Since the ladder's length

ℓ

is constant, the key is to express the horizontal and vertical positions of the ladder's ends in terms of

ℓ

and the angles

α

and

β

. Don't overcomplicate; just draw the two triangles.

Hint 2

Once you have the expressions for the horizontal shift

a

and the vertical shift

b

, you will be left with a ratio involving differences of trigonometric functions. This is a classic signal to deploy the Sum-to-Product identities. Specifically, look for

cos β - cos α

and

sin α - sin β

Hint 3

Start by writing the ratio

\frac{a}{b} = \frac{ℓ ( c o s β - c o s α )}{ℓ ( s i n α - s i n β )}

. Once you cancel

ℓ

, apply the identity

cos C - cos D = 2 sin (\frac{C + D}{2}) sin (\frac{D - C}{2})

to the numerator and

sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})

to the denominator.

Visualized Solution

Initial Position of the Ladder

Let the length of the ladder be $ℓ$ .
The ladder rests against the wall, making an angle $α$ with the horizontal floor.
The foot is at point $A$ and the top is at point $B$ .

Final Position of the Ladder

The foot of the ladder is pulled away from the wall.
The new position of the foot is $A^{'}$ and the top slides down to $B^{'}$ .
The new angle with the horizontal is $β$ .

Coordinates of Initial Position

Using right-angled triangle $△ O A B$ :
Distance of foot from wall: $O A = ℓ cos α$
Height of top from floor: $O B = ℓ sin α$

Coordinates of Final Position

Using right-angled triangle $△ O A^{'} B^{'}$ :
New distance of foot from wall: $O A^{'} = ℓ cos β$
New height of top from floor: $O B^{'} = ℓ sin β$

Calculating Horizontal Shift $a$

The foot is pulled away by a distance $a$ .
$a = O A^{'} - O A$
$a = ℓ cos β - ℓ cos α$
$a = ℓ (cos β - cos α)$

Calculating Vertical Shift $b$

The top slides down by a distance $b$ .
$b = O B - O B^{'}$
$b = ℓ sin α - ℓ sin β$
$b = ℓ (sin α - sin β)$

Forming the Ratio $\frac{a}{b}$

We need to prove a relation between $a$ and $b$ . Let's divide $a$ by $b$ :
$\frac{a}{b} = \frac{ℓ ( c o s β - c o s α )}{ℓ ( s i n α - s i n β )}$
Canceling $ℓ$ :
$\frac{a}{b} = \frac{c o s β - c o s α}{s i n α - s i n β}$

Trigonometric Transformations

We need the Sum-to-Product formulas to simplify the expression.
For the numerator: $cos C - cos D = 2 sin (\frac{C + D}{2}) sin (\frac{D - C}{2})$
For the denominator: $sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})$

Simplifying the Numerator

Apply $cos C - cos D$ to $cos β - cos α$ :
Here, $C = β$ and $D = α$ .
Numerator: $2 sin (\frac{β + α}{2}) sin (\frac{α - β}{2})$
Note the order in the second sine term: $\frac{D - C}{2} \to \frac{α - β}{2}$

Simplifying the Denominator

Apply $sin C - sin D$ to $sin α - sin β$ :
Here, $C = α$ and $D = β$ .
Denominator: $2 cos (\frac{α + β}{2}) sin (\frac{α - β}{2})$

Substituting and Canceling Terms

Substitute back into the ratio:
$\frac{a}{b} = \frac{2 s i n ( \frac{α + β}{2} ) s i n ( \frac{α - β}{2} )}{2 c o s ( \frac{α + β}{2} ) s i n ( \frac{α - β}{2} )}$
Cancel the common factor $2$ .
Cancel the common factor $sin (\frac{α - β}{2})$ .
$\frac{a}{b} = \frac{s i n ( \frac{α + β}{2} )}{c o s ( \frac{α + β}{2} )}$

Final Result

We know that $\frac{s i n θ}{c o s θ} = tan θ$ .
Therefore, $\frac{a}{b} = tan (\frac{α + β}{2})$
Rearranging the terms:
$a = b tan (\frac{α + β}{2})$
Hence Proved.

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine a ladder of constant length

ℓ

leaning against a wall. It rests at an angle

α

with the floor, with the foot at point

A

and the top at point

B

By dropping a perpendicular from the ladder to the floor, we create a right-angled triangle. The horizontal distance from the wall is

O A = ℓ cos α

, and the vertical height is

O B = ℓ sin α

The Shift

Defining the Change

When the foot is pulled away to a new position

A^{'}

, the ladder slides such that the top descends to

B^{'}

. The ladder maintains its length

ℓ

, but the angle changes to

β

The new horizontal distance is

O A^{'} = ℓ cos β

, and the new vertical height is

O B^{'} = ℓ sin β

. We define the horizontal shift

a

and the vertical shift

b

as follows:

a = O A^{'} - O A = ℓ (cos β - cos α)

b = O B - O B^{'} = ℓ (sin α - sin β)

The Trigonometric Dance

To prove the relationship

a = b tan \frac{1}{2} (α + β)

, we must eliminate the length

ℓ

. We take the ratio of the two shifts:

\frac{a}{b} = \frac{ℓ ( cos β - cos α )}{ℓ ( sin α - sin β )} = \frac{cos β - cos α}{sin α - sin β}

To simplify this, we apply the Sum-to-Product identities. For the numerator, we use

cos C - cos D = 2 sin (\frac{C + D}{2}) sin (\frac{D - C}{2})

. For the denominator, we use

sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})

Applying these identities yields:

\frac{a}{b} = \frac{2 sin ( \frac{β + α}{2} ) sin ( \frac{α - β}{2} )}{2 cos ( \frac{α + β}{2} ) sin ( \frac{α - β}{2} )}

The Grand Finale

Observe that the factor

2

and the term

sin (\frac{α - β}{2})

appear in both the numerator and the denominator. Canceling these common terms leaves us with:

\frac{a}{b} = \frac{sin ( \frac{α + β}{2} )}{cos ( \frac{α + β}{2} )}

Recognizing the definition of the tangent function, we obtain:

\frac{a}{b} = tan (\frac{α + β}{2})

This confirms the final relationship:

$a = b tan (\frac{α + β}{2})$

Common Traps & FAQs

Wait, does the length of the ladder

ℓ

actually matter? If it cancels out, does that mean any ladder would behave the same way?

That is a brilliant observation! Yes, the length

ℓ

cancels out because the relationship between the shifts

a

and

b

is purely geometric—it depends only on the angular change. It implies that the 'geometry of the slide' is scale-invariant. Whether your ladder is 2 meters or 20 meters, the ratio of the shifts depends solely on the angles.

I always mix up the signs in the Sum-to-Product formulas. Is there a trick?

The struggle is real! The best way is to derive them from the compound angle formulas (like

sin (A + B)

and

sin (A - B)

) once. Once you see the pattern—that sine differences involve a cosine term and cosine differences involve sine terms—it becomes intuitive. Also, remember that

cos C - cos D

is the 'odd one out' because it produces a negative sign or requires swapping

D - C

to keep it clean.

Can we solve this using coordinate geometry? Let the wall be the y-axis and the floor the x-axis?

Absolutely! You can set the coordinates of the foot as

(x, 0)

and the top as

(0, y)

. Then

x^{2} + y^{2} = ℓ^{2}

. You can differentiate or use the change in coordinates. However, the trigonometric approach is often faster and more elegant for this specific problem. In JEE, choosing the path of least resistance is a skill in itself!

Visualized Solution

Initial Position of the Ladder

Final Position of the Ladder

Coordinates of Initial Position

Coordinates of Final Position

Calculating Horizontal Shift $a$

Calculating Vertical Shift $b$

Forming the Ratio $\frac{a}{b}$

Trigonometric Transformations

Simplifying the Numerator

Simplifying the Denominator

Substituting and Canceling Terms

Final Result

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Shift

The Trigonometric Dance

The Grand Finale

Similar Questions

A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is

A vertical pole stands at a point $Q$ on a horizontal ground. $A$ and $B$ are points on the ground, $d$ meters apart. The pole subtends angles $α$ and $β$ at $A$ and $B$ respectively. $A B$ subtends an angle $γ$ at $Q$ . Find the height of the pole.

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If $∠ B P C = β$ , then $tan β$ is equal to:

The angle of elevation of the top of a tower from a point $A$ due north of it is $α$ and from a point $B$ at a distance of $9$ units due west of $A$ is $cos^{- 1} (\frac{3}{13})$ . If the distance of the point $B$ from the tower is $15$ units, then $cot α$ is equal to :

Visualized Solution

Initial Position of the Ladder

Final Position of the Ladder

Coordinates of Initial Position

Coordinates of Final Position

Calculating Horizontal Shift .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Calculating Vertical Shift .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Forming the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ba​

Trigonometric Transformations

Simplifying the Numerator

Simplifying the Denominator

Substituting and Canceling Terms

Final Result

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Shift

The Trigonometric Dance

The Grand Finale

Similar Questions

A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is

Calculating Horizontal Shift $a$

Calculating Vertical Shift $b$

Forming the Ratio $\frac{a}{b}$