Why is the perpendicular distance $OQ = a \sin(\frac{\alpha}{2})$?

Look at the right-angled triangle $\triangle OQP$. The hypotenuse is $OP$, which is the distance $a$. The angle at vertex $P$ is the angle of rotation, which is $\frac{\alpha}{2}$. By definition, $\sin(\text{angle}) = \frac{\text{Opposite}}{\text{Hypotenuse}}$. Here, the opposite side is the perpendicular distance $OQ$. Thus, $\sin(\frac{\alpha}{2}) = \frac{OQ}{a}$, which rearranges to $OQ = a \sin(\frac{\alpha}{2})$.

JEE Main 2025 (April)

LEVELJEE Advanced

Animated Solution for Mathematics - Straight Lines: A line passing through the point $P (a, θ)$ makes an acute angle $α$ with the positive $x$ -axis. Let this line be rotated about the point $P$ through an angle $\frac{α}{2}$ in the clock-wise direction. If in the new position, the slope of the line is $2 - 3$ and its distance from the origin is $\frac{1}{2}$ , then the value of $3 a^{2} tan^{2} α - 23$ is

Q: Wait, why can we just assume $\theta = \alpha$?

Great question! The problem defines $P$ in polar coordinates $(a, \theta)$ and states the line passes through $P$ with inclination $\alpha$. Since the choice of the coordinate system's orientation is arbitrary, we can align the line such that the segment $OP$ lies on the line itself. This is a powerful technique in coordinate geometry to simplify calculations without losing generality.

Select Answer:

Visualized Solution

Initial Setup for $P (a, θ)$

Let the original line pass through $P (a, θ)$ and make an angle $α$ with the $x$ -axis.
For a consistent geometric configuration, we assume $θ = α$ .
This implies the original line passes directly through the origin $O (0, 0)$ .
The distance $O P$ is exactly $a$ .

Rotating the Line by $\frac{α}{2}$

The line is rotated clockwise about point $P$ by an angle of $\frac{α}{2}$ .
The new inclination of the line becomes $α - \frac{α}{2} = \frac{α}{2}$ .

Finding the Angle $α$

The slope of the new rotated line is given as $2 - 3$ .
Therefore, $tan (\frac{α}{2}) = 2 - 3$ .
From standard trigonometric values, we know $tan (1 5^{\circ}) = 2 - 3$ .
Equating the angles: $\frac{α}{2} = 1 5^{\circ} ⟹ α = 3 0^{\circ}$ .

The Perpendicular Distance $d$

The perpendicular distance from the origin $O$ to the rotated line is $d = \frac{1}{2}$ .
Let $Q$ be the foot of the perpendicular from $O$ to the rotated line.
This forms a right-angled triangle, $△ O QP$ .

Trigonometry in $△ O QP$

In $△ O QP$ , the hypotenuse is $O P = a$ .
The angle $∠ O P Q$ is exactly the angle of rotation, which is $\frac{α}{2} = 1 5^{\circ}$ .
Using sine ratio: $sin (1 5^{\circ}) = \frac{Opposite}{Hypotenuse} = \frac{O Q}{O P}$ .
Therefore, $O Q = a sin (1 5^{\circ})$ .

Equating the Distance to $\frac{1}{2}$

We are given that the distance $O Q = \frac{1}{2}$ .
Substituting our expression: $a sin (1 5^{\circ}) = \frac{1}{2}$ .
To eliminate the square root and prepare for finding $a^{2}$ , we square both sides.
$a^{2} sin^{2} (1 5^{\circ}) = \frac{1}{2}$ .

Calculating $sin^{2} (1 5^{\circ})$

We need the exact value of $sin^{2} (1 5^{\circ})$ .
Using the half-angle identity: $sin^{2} (θ) = \frac{1 - c o s ( 2 θ )}{2}$ .
$sin^{2} (1 5^{\circ}) = \frac{1 - c o s ( 3 0 ^{\circ} )}{2}$ .
Substituting $cos (3 0^{\circ}) = \frac{3}{2}$ , we get $sin^{2} (1 5^{\circ}) = \frac{1 - \frac{3}{2}}{2} = \frac{2 - 3}{4}$ .

Solving for $a^{2}$

Substitute $sin^{2} (1 5^{\circ})$ back into our equation: $a^{2} (\frac{2 - 3}{4}) = \frac{1}{2}$ .
Rearranging for $a^{2}$ : $a^{2} = \frac{1}{2} \times \frac{4}{2 - 3} = \frac{2}{2 - 3}$ .
Rationalizing the denominator by multiplying by $\frac{2 + 3}{2 + 3}$ :
$a^{2} = 2 (2 + 3) = 4 + 23$ .

Setting Up $3 a^{2} tan^{2} α - 23$

The problem asks for the value of: $3 a^{2} tan^{2} α - 23$ .
We already found $α = 3 0^{\circ}$ , so $tan α = \frac{1}{3}$ .
Squaring this gives $tan^{2} α = \frac{1}{3}$ .
We also have $a^{2} = 4 + 23$ .

Final Calculation to get $4$

Substitute the values into the expression:
$3 (4 + 23) (\frac{1}{3}) - 23$ .
The $3$ and $\frac{1}{3}$ cancel each other out perfectly.
We are left with: $4 + 23 - 23$ .
The $+ 23$ and $- 23$ cancel out, leaving exactly $4$ .

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The Sigma Insight: Distance of a Point from a Line

Solution Diagram

Analyzing the Setup

Imagine standing on a coordinate plane, looking at a line passing through a point

P (a, α)

. By setting the inclination to

α

, we align our line such that it passes through the origin

O (0, 0)

The segment

O P

acts as a vector with length

a

. This serves as our primary reference for the geometric transformation.

The Clockwise Swing

The line rotates clockwise about

P

by an angle of

\frac{α}{2}

. Given the original inclination was

α

, the new inclination becomes:

α - \frac{α}{2} = \frac{α}{2}

The slope of this new line is

2 - 3

. Since the slope is the tangent of the inclination, we have:

tan (\frac{α}{2}) = 2 - 3

Recognizing this as

tan (1 5^{\circ})

is the critical step. Thus,

\frac{α}{2} = 1 5^{\circ}

, which implies

α = 3 0^{\circ}

The Hidden Triangle

Let

Q

be the foot of the perpendicular from the origin to the new line. We form a right-angled triangle

△ O QP

where the hypotenuse

O P = a

and the angle at

P

1 5^{\circ}

Using the sine ratio, the perpendicular distance

O Q

is:

O Q = a sin (1 5^{\circ})

Given

O Q = \frac{1}{2}

, we equate the expressions:

a sin (1 5^{\circ}) = \frac{1}{2}

Squaring both sides yields:

a^{2} sin^{2} (1 5^{\circ}) = \frac{1}{2}

The Final Elegance

Using the identity

sin^{2} (θ) = \frac{1 - c o s ( 2 θ )}{2}

, we calculate:

sin^{2} (1 5^{\circ}) = \frac{1 - cos ( 3 0 ^{\circ} )}{2} = \frac{1 - \frac{3}{2}}{2} = \frac{2 - 3}{4}

Substituting this into our equation for

a^{2}

a^{2} (\frac{2 - 3}{4}) = \frac{1}{2} \Rightarrow a^{2} = \frac{2}{2 - 3}

Rationalizing the denominator, we find:

a^{2} = 2 (2 + 3) = 4 + 23

We now evaluate the expression

3 a^{2} tan^{2} α - 23

. With

α = 3 0^{\circ}

, we know

tan^{2} (3 0^{\circ}) = \frac{1}{3}

Substituting these values:

3 (4 + 23) (\frac{1}{3}) - 23 = 4 + 23 - 23 = 4

The final result is 4.

Common Traps & FAQs

Wait, why can we just assume

θ = α

Great question! The problem defines

P

in polar coordinates

(a, θ)

and states the line passes through

P

with inclination

α

. Since the choice of the coordinate system's orientation is arbitrary, we can align the line such that the segment

O P

lies on the line itself. This is a powerful technique in coordinate geometry to simplify calculations without losing generality.

How do I know

tan (1 5^{\circ}) = 2 - 3

You can derive this using the half-angle formula or the subtraction formula:

tan (1 5^{\circ}) = tan (4 5^{\circ} - 3 0^{\circ}) = \frac{t a n 4 5 ^{\circ} - t a n 3 0 ^{\circ}}{1 + t a n 4 5 ^{\circ} t a n 3 0 ^{\circ}}

. Substituting the values

tan 4 5^{\circ} = 1

and

tan 3 0^{\circ} = \frac{1}{3}

, you get

\frac{1 - 1/ 3}{1 + 1/ 3} = \frac{3 - 1}{3 + 1}

. Rationalizing this gives

2 - 3

Why is the perpendicular distance

O Q = a sin (\frac{α}{2})

Look at the right-angled triangle

△ O QP

. The hypotenuse is

O P

, which is the distance

a

. The angle at vertex

P

is the angle of rotation, which is

\frac{α}{2}

. By definition,

sin (angle) = \frac{Opposite}{Hypotenuse}

. Here, the opposite side is the perpendicular distance

O Q

. Thus,

sin (\frac{α}{2}) = \frac{O Q}{a}

, which rearranges to

O Q = a sin (\frac{α}{2})

Select Answer:

Visualized Solution

Initial Setup for $P (a, θ)$

Rotating the Line by $\frac{α}{2}$

Finding the Angle $α$

The Perpendicular Distance $d$

Trigonometry in $△ O QP$

Equating the Distance to $\frac{1}{2}$

Calculating $sin^{2} (1 5^{\circ})$

Solving for $a^{2}$

Setting Up $3 a^{2} tan^{2} α - 23$

Final Calculation to get $4$

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Clockwise Swing

The Hidden Triangle

The Final Elegance

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Visualized Solution

Initial Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(a,θ)

Rotating the Line by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2α​

Finding the Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

The Perpendicular Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Trigonometry in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OQP

Equating the Distance to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2​1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2(15∘)

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

Setting Up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3a2tan2α−23​

Final Calculation to get .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Clockwise Swing

The Hidden Triangle

The Final Elegance

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Rotating the Line by $\frac{α}{2}$

Finding the Angle $α$

The Perpendicular Distance $d$

Trigonometry in $△ O QP$

Equating the Distance to $\frac{1}{2}$

Calculating $sin^{2} (1 5^{\circ})$

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Setting Up $3 a^{2} tan^{2} α - 23$

Final Calculation to get $4$

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