JEE Advanced 1991

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: In a triangle of base $a$ the ratio of the other two sides is $r (< 1)$ . Show that the altitude of the triangle is less than or equal to $\frac{a r}{1 - r ^{2}}$ .

Q: Why did we multiply by $\sin(B-C)$? It feels like pulling a rabbit out of a hat!

It is a strategic move! We need to create a difference of squares in the denominator to match the $1-r^2$ structure. By multiplying the numerator and denominator by $\sin(B-C)$, we invoke the identity $\sin(B+C)\sin(B-C) = \sin^2 B - \sin^2 C$, which perfectly sets the stage for converting sines into side lengths.

Q: Can't we use the Area formula $Area = \frac{1}{2}ah$?

You absolutely can! The area of a triangle is $\frac{1}{2}ah$, but also $\frac{1}{2}bc \sin A$. Equating these gives $h = \frac{bc \sin A}{a}$. This is a valid path, but the Sine Rule approach we used is often more direct when dealing with ratios of sides. Both roads lead to Rome, but one might be smoother!

Q: Why is $r < 1$ given?

That is a crucial constraint! It ensures that the ratio $r = \frac{c}{b}$ is less than 1, which guarantees that $1-r^2$ is positive. If $r$ were greater than 1, the inequality would flip, and the geometry would change. Always respect the constraints given in the problem statement; they are the guardrails of your derivation.

Visualized Solution

Visualizing Triangle $A B C$ and Altitude $h$

Consider $Δ A B C$ with base $B C = a$ .
Let the altitude from $A$ to $B C$ be $h$ .

Expressing Altitude using Trigonometry

In the right-angled $Δ A B D$ , we have $sin B = \frac{h}{c}$ .
Therefore, the altitude can be expressed as: $h = c sin B$ .

Applying the Sine Rule

By the Sine Rule: $\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} = k$ .
From this, we can write: $a = k sin A$ , $b = k sin B$ , and $c = k sin C$ .

Relating $h$ to Sides $a, b, c$

Start with $h = c sin B$ .
Multiply and divide by $a$ : $h = \frac{c a s i n B}{a}$ .
Substitute $a = k sin A$ in the denominator: $h = \frac{c a s i n B}{k s i n A}$ .

Introducing $sin (B - C)$

Multiply numerator and denominator by $sin (B - C)$ .
$h = \frac{c a s i n B s i n ( B - C )}{k s i n A s i n ( B - C )}$ .

Using Trigonometric Identities

Since $A + B + C = π$ , $sin A = sin (B + C)$ .
The denominator becomes $k sin (B + C) sin (B - C)$ .
Using the identity: $sin (B + C) sin (B - C) = sin^{2} B - sin^{2} C$ .

Converting Sines back to Sides

Denominator: $k (sin^{2} B - sin^{2} C)$ .
From Sine Rule, $sin B = \frac{b}{k}$ and $sin C = \frac{c}{k}$ .
Denominator becomes $k (\frac{b ^{2}}{k ^{2}} - \frac{c ^{2}}{k ^{2}}) = \frac{b ^{2} - c ^{2}}{k}$ .

Simplifying the Expression for $h$

Substitute the denominator back: $h = \frac{c a s i n B s i n ( B - C )}{( b ^{2} - c ^{2} ) / k}$ .
Rearrange: $h = \frac{a c ( k s i n B ) s i n ( B - C )}{b ^{2} - c ^{2}}$ .
Since $k sin B = b$ , we get: $h = \frac{ab c s i n ( B - C )}{b ^{2} - c ^{2}}$ .

Introducing the Ratio $r$

We are given the ratio of the other two sides: $r = \frac{c}{b}$ .
Divide the numerator and denominator of $h$ by $b^{2}$ .
$h = \frac{a ( \frac{c}{b} ) s i n ( B - C )}{1 - ( \frac{c}{b} ) ^{2}}$ .

Substituting $r$ into the Equation

Substitute $\frac{c}{b} = r$ into the expression.
$h = \frac{a r s i n ( B - C )}{1 - r ^{2}}$ .

Establishing the Maximum Altitude

For any angle $θ$ , the maximum value of $sin θ$ is $1$ .
Therefore, $sin (B - C) \leq 1$ .
Applying this: $h = \frac{a r s i n ( B - C )}{1 - r ^{2}} \leq \frac{a r}{1 - r ^{2}}$ .
Hence, $h \leq \frac{a r}{1 - r ^{2}}$ . Proved.

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The Sigma Insight: Properties of Triangles

Solution Diagram

Analyzing the Setup

We begin with

Δ A B C

having a base

B C = a

. We drop a perpendicular from vertex

A

to the base

B C

, intersecting at point

D

. This segment

A D

represents our altitude

h

In the right-angled triangle

Δ A B D

, we observe the fundamental trigonometric relationship:

sin B = \frac{h}{c}

This yields our primary anchor equation:

h = c sin B

The Sine Rule Bridge

To connect the altitude to the broader geometry of the triangle, we employ the Sine Rule:

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = k

This allows us to express any side in terms of its opposite angle. To introduce the base

a

into our expression for

h

, we manipulate the equation as follows:

h = \frac{c a sin B}{a}

Substituting

a = k sin A

into the expression, we obtain:

h = \frac{c a sin B}{k sin A}

The Algebraic Alchemy

We now aim to create a denominator that reflects the structure

1 - r^{2}

. Given that

A + B + C = π

, we know that

sin A = sin (B + C)

By multiplying both the numerator and denominator by

sin (B - C)

, we utilize the trigonometric identity:

sin (B + C) sin (B - C) = sin^{2} B - sin^{2} C

This transforms our denominator into

k (sin^{2} B - sin^{2} C)

. Applying the Sine Rule again, where

sin B = \frac{b}{k}

and

sin C = \frac{c}{k}

, the denominator simplifies to:

\frac{b ^{2} - c ^{2}}{k}

Substituting this back into our expression for

h

, the constant

k

cancels out, leaving us with:

h = \frac{ab c sin ( B - C )}{b ^{2} - c ^{2}}

The Final Reveal

We are given the ratio

r = \frac{c}{b} < 1

. To express

h

in terms of

r

, we divide the numerator and denominator by

b^{2}

h = \frac{a ( \frac{c}{b} ) sin ( B - C )}{1 - ( \frac{c}{b} ) ^{2}}

This simplifies to the elegant form:

h = \frac{a r sin ( B - C )}{1 - r ^{2}}

Since the maximum value of

sin (B - C)

1

, we conclude that the altitude is bounded by: $h \leq \frac{a r}{1 - r ^{2}}$

We have successfully arrived at our destination. In JEE Advanced, the elegance of the derivation is as critical as the final result.

Common Traps & FAQs

Why did we multiply by

sin (B - C)

? It feels like pulling a rabbit out of a hat!

It is a strategic move! We need to create a difference of squares in the denominator to match the

1 - r^{2}

structure. By multiplying the numerator and denominator by

sin (B - C)

, we invoke the identity

sin (B + C) sin (B - C) = sin^{2} B - sin^{2} C

, which perfectly sets the stage for converting sines into side lengths.

Can't we use the Area formula

A r e a = \frac{1}{2} ah

You absolutely can! The area of a triangle is

\frac{1}{2} ah

, but also

\frac{1}{2} b c sin A

. Equating these gives

h = \frac{b c s i n A}{a}

. This is a valid path, but the Sine Rule approach we used is often more direct when dealing with ratios of sides. Both roads lead to Rome, but one might be smoother!

Why is

r < 1

given?

That is a crucial constraint! It ensures that the ratio

r = \frac{c}{b}

is less than 1, which guarantees that

1 - r^{2}

is positive. If

r

were greater than 1, the inequality would flip, and the geometry would change. Always respect the constraints given in the problem statement; they are the guardrails of your derivation.

Animated Solution for Mathematics - Trigonometry: In a triangle of base $a$ the ratio of the other two sides is $r (< 1)$ . Show that the altitude of the triangle is less than or equal to $\frac{a r}{1 - r ^{2}}$ .

Visualized Solution

Visualizing Triangle $A B C$ and Altitude $h$

Expressing Altitude using Trigonometry

Applying the Sine Rule

Relating $h$ to Sides $a, b, c$

Introducing $sin (B - C)$

Using Trigonometric Identities

Converting Sines back to Sides

Simplifying the Expression for $h$

Introducing the Ratio $r$

Substituting $r$ into the Equation

Establishing the Maximum Altitude

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Sine Rule Bridge

The Algebraic Alchemy

The Final Reveal

Similar Questions

If $Δ$ is the area of a triangle with side lengths $a, b, c$ , then show that $Δ \leq \frac{1}{4} (a + b + c) ab c$ . Also show that the equality occurs in the above inequality if and only if $a = b = c$ .

Consider the following statements concerning a triangle $A B C$ : (i) The sides $a, b, c$ and area $Δ$ are rational. (ii) $a, tan \frac{B}{2}, tan \frac{C}{2}$ are rational. (iii) $a, sin A, sin B, sin C$ are rational. Prove that (i) $⟹$ (ii) $⟹$ (iii) $⟹$ (i).

If in a triangle $A B C$ , $cos A cos B + sin A sin B sin C = 1$ , Show that $a : b : c = 1 : 1 : 2$ .

There exists a triangle $A B C$ satisfying the conditions

A circle is inscribed in an equilateral triangle of side $a$ . The area of any square inscribed in this circle is ..................

Consider a triangle $P QR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$ , respectively. Then which of the following statements is (are) TRUE?

In a $Δ A B C$ , $∠ A = 9 0^{\circ}$ and $A D$ is an altitude. Complete the relation $\frac{B D}{B A} = \frac{A B}{( \dots )}$

Visualized Solution

Visualizing Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC and Altitude .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

Expressing Altitude using Trigonometry

Applying the Sine Rule

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h to Sides .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,c

Introducing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(B−C)

Using Trigonometric Identities

Converting Sines back to Sides

Simplifying the Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

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

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r into the Equation

Establishing the Maximum Altitude

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Sine Rule Bridge

The Algebraic Alchemy

The Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If in a triangle ABC, cosAcosB+sinAsinBsinC=1, Show that a:b:c=1:1:2​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } There exists a triangle ABC satisfying the conditions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle is inscribed in an equilateral triangle of side a. The area of any square inscribed in this circle is ..................

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider a triangle PQR having sides of lengths p,q and r opposite to the angles P,Q and R, respectively. Then which of the following statements is (are) TRUE?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a ΔABC, ∠A=90∘ and AD is an altitude. Complete the relation BABD​=(…)AB​

Visualizing Triangle $A B C$ and Altitude $h$

Relating $h$ to Sides $a, b, c$

Introducing $sin (B - C)$

Simplifying the Expression for $h$

Introducing the Ratio $r$

Substituting $r$ into the Equation

If in a triangle $A B C$ , $cos A cos B + sin A sin B sin C = 1$ , Show that $a : b : c = 1 : 1 : 2$ .

There exists a triangle $A B C$ satisfying the conditions

A circle is inscribed in an equilateral triangle of side $a$ . The area of any square inscribed in this circle is ..................

Consider a triangle $P QR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$ , respectively. Then which of the following statements is (are) TRUE?

In a $Δ A B C$ , $∠ A = 9 0^{\circ}$ and $A D$ is an altitude. Complete the relation $\frac{B D}{B A} = \frac{A B}{( \dots )}$