JEE Main 2019 (11 January)

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: In a triangle, the sum of lengths of two sides is $x$ and the product of the lengths of the same two sides is $y$ . If $x^{2} - c^{2} = y$ , where $c$ is the length of the third side of the triangle, then the circumradius of the triangle is :

Q: Wait, why is $\cos C = -\frac{1}{2}$ implying $C = 120^\circ$ and not something else?

In a triangle, the sum of angles is $180^\circ$. Since $\cos C$ is negative, $C$ must be obtuse (between $90^\circ$ and $180^\circ$). The only angle in this range with a cosine of $-\frac{1}{2}$ is $120^\circ$. It is a beautiful, unique solution!

Q: Why can't we just use Heron's formula to find the area and then use $R = \frac{abc}{4\Delta}$?

You absolutely could! But that would be a long, winding road. In JEE, time is your most precious resource. The Sine Rule is a direct shortcut. Always look for the path of least resistance.

Q: Is it possible for the sides $a$ and $b$ to be anything? Does the triangle exist?

Great question! The triangle inequality must hold ($a+b > c$, etc.). Since we found a valid angle $C=120^\circ$, such a triangle definitely exists for appropriate values of $a$ and $b$. The math confirms the geometry!

Select Answer:

Visualized Solution

Visualizing the Triangle

Let the sides of the triangle be $a$ , $b$ , and $c$ .
Let the angle opposite to side $c$ be $C$ .

Defining $x$ and $y$

Sum of two sides: $a + b = x$
Product of two sides: $ab = y$

Substitution in the Given Relation

Given: $x^{2} - c^{2} = y$
Substitute $x = a + b$ and $y = ab$ :
$(a + b)^{2} - c^{2} = ab$

Expanding the Square

Expand $(a + b)^{2}$ :
$a^{2} + b^{2} + 2 ab - c^{2} = ab$

Simplifying the Equation

Rearrange the terms:
$a^{2} + b^{2} - c^{2} = ab - 2 ab$
$a^{2} + b^{2} - c^{2} = - ab$

Applying the Cosine Rule

Recall the Cosine Rule:
$cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$

Solving for $cos C$

Substitute $a^{2} + b^{2} - c^{2} = - ab$ :
$cos C = \frac{- ab}{2 ab}$
$cos C = - \frac{1}{2}$

Finding Angle $C$

Since $cos C = - \frac{1}{2}$ and $0^{\circ} < C < 18 0^{\circ}$ :
$C = 12 0^{\circ}$

Introducing the Sine Rule

The formula for circumradius $R$ is:
$R = \frac{c}{2 s i n C}$

Calculating $sin 12 0^{\circ}$

Calculate $sin 12 0^{\circ}$ :
$sin 12 0^{\circ} = sin (18 0^{\circ} - 6 0^{\circ})$
$sin 12 0^{\circ} = \frac{3}{2}$

Final Computation of $R$

Substitute $sin 12 0^{\circ}$ into the formula for $R$ :
$R = \frac{c}{2 ( \frac{3}{2} )}$
$R = \frac{c}{3}$

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The Sigma Insight: Properties of Triangles

Solution Diagram

Analyzing the Setup

Imagine you are standing on the edge of a triangle with sides

a

b

, and

c

. We are given that the sum of two sides is

x

and their product is

y

, satisfying the relation

x^{2} - c^{2} = y

Our goal is to determine the circumradius

R

of this triangle.

Decoding the Cipher

First, we translate the problem into the language of algebra. We have the definitions

a + b = x

and

ab = y

Substituting these into the given relation

x^{2} - c^{2} = y

, we obtain:

(a + b)^{2} - c^{2} = ab

Expanding the square, we get

a^{2} + b^{2} + 2 ab - c^{2} = ab

. Rearranging the terms by moving

2 ab

to the right side yields:

a^{2} + b^{2} - c^{2} = - ab

The Geometric Revelation

Whenever you encounter the expression

a^{2} + b^{2} - c^{2}

in a triangle, the Cosine Rule is the natural tool to employ. The rule states:

cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}

Substituting our derived relation

a^{2} + b^{2} - c^{2} = - ab

into the Cosine Rule, we find:

cos C = \frac{- ab}{2 ab} = - \frac{1}{2}

Since

C

is an interior angle of a triangle,

cos C = - \frac{1}{2}

implies that $C = 12 0^{\circ}$ . We have successfully identified that the triangle is obtuse.

The Final Leap

Now that we have determined the angle

C

, we calculate the circumradius

R

using the Extended Sine Rule:

R = \frac{c}{2 sin C}

We know that

sin 12 0^{\circ} = sin (18 0^{\circ} - 6 0^{\circ}) = sin 6 0^{\circ} = \frac{3}{2}

. Substituting this value into the formula gives:

R = \frac{c}{2 ( \frac{3}{2} )}

The factors of

2

cancel out, leading us to the final, elegant result:

$R = \frac{c}{3}$

Common Traps & FAQs

Wait, why is

cos C = - \frac{1}{2}

implying

C = 12 0^{\circ}

and not something else?

In a triangle, the sum of angles is

18 0^{\circ}

. Since

cos C

is negative,

C

must be obtuse (between

9 0^{\circ}

and

18 0^{\circ}

). The only angle in this range with a cosine of

- \frac{1}{2}

12 0^{\circ}

. It is a beautiful, unique solution!

Why can't we just use Heron's formula to find the area and then use

R = \frac{ab c}{4Δ}

You absolutely could! But that would be a long, winding road. In JEE, time is your most precious resource. The Sine Rule is a direct shortcut. Always look for the path of least resistance.

Is it possible for the sides

a

and

b

to be anything? Does the triangle exist?

Great question! The triangle inequality must hold (

a + b > c

, etc.). Since we found a valid angle

C = 12 0^{\circ}

, such a triangle definitely exists for appropriate values of

a

and

b

. The math confirms the geometry!

Animated Solution for Mathematics - Trigonometry: In a triangle, the sum of lengths of two sides is $x$ and the product of the lengths of the same two sides is $y$ . If $x^{2} - c^{2} = y$ , where $c$ is the length of the third side of the triangle, then the circumradius of the triangle is :

Select Answer:

Visualized Solution

Visualizing the Triangle

Defining $x$ and $y$

Substitution in the Given Relation

Expanding the Square

Simplifying the Equation

Applying the Cosine Rule

Solving for $cos C$

Finding Angle $C$

Introducing the Sine Rule

Calculating $sin 12 0^{\circ}$

Final Computation of $R$

The Sigma Insight: Properties of Triangles

Analyzing the Setup

Decoding the Cipher

The Geometric Revelation

The Final Leap

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Triangle

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Substitution in the Given Relation

Expanding the Square

Simplifying the Equation

Applying the Cosine Rule

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

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

Introducing the Sine Rule

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin120∘

Final Computation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

The Sigma Insight: Properties of Triangles

Analyzing the Setup

Decoding the Cipher

The Geometric Revelation

The Final Leap

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