JEE Advanced 1980

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: $A B C$ is a triangle. $D$ is the middle point of $B C$ . If $A D$ is perpendicular to $A C$ , then prove that $cos A cos C = \frac{2 ( c ^{2} - a ^{2} )}{3 a c}$ .

Q: Is there a way to solve this using vectors?

Yes, absolutely! You could place vertex $A$ at the origin $(0,0)$ and define the coordinates of $B$ and $C$. Since $AD \perp AC$, you could use the dot product to represent the perpendicularity condition. While that is a powerful method, the geometric/trigonometric approach we used is often preferred in JEE for its elegance and speed. Both methods are valid, but mastering this trigonometric path will sharpen your intuition for similar problems.

Progressive Hints

Hint 1

Hint 1: The Spark. Focus on the geometric constraint given:

A D ⊥ A C

. This creates a right-angled triangle

Δ A C D

. In any right-angled triangle, the cosine of an angle is simply the ratio of the adjacent side to the hypotenuse. How can you express

cos C

using only the sides

A C

and

C D

Hint 2

Hint 2: The Arsenal. Remember that the Cosine Rule is your most powerful tool for relating sides and angles in any triangle. For

Δ A B C

, the Cosine Rule states

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

. You now have two different expressions for

cos C

. Equating them will allow you to find a crucial relationship between

a

b

, and

c

Hint 3

Hint 3: The Execution. Once you equate the two expressions for

cos C

, you will find that

b^{2} = \frac{a ^{2} - c ^{2}}{3}

. Now, write the expression for

cos A

using the Cosine Rule:

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

. Multiply this by your expression for

cos C

and substitute the value of

b^{2}

to reach the final proof.

Visualized Solution

$V i s u a l i z in g t h e T r ian g l e A B C$

Given triangle $A B C$ with $D$ as the midpoint of $B C$ .
Therefore, $B D = D C = \frac{a}{2}$ .
It is given that $A D ⊥ A C$ , so $∠ D A C = 9 0^{\circ}$ .

$T r i g o n o m e t r y in Δ A C D$

In right-angled $Δ A C D$ :
$cos C = \frac{Adjacent}{Hypotenuse} = \frac{A C}{C D}$
Substitute $A C = b$ and $C D = \frac{a}{2}$ :
$cos C = \frac{b}{a /2} = \frac{2 b}{a}$

$C os in e R u l e f or Δ A B C$

In $Δ A B C$ , using the Cosine Rule for $∠ C$ :
$cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$

$E q u a t in g t h e E x p r ess i o n s$

Equating the two expressions for $cos C$ :
$\frac{2 b}{a} = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$
Multiply both sides by $2 ab$ :
$4 b^{2} = a^{2} + b^{2} - c^{2}$

$S o l v in g f or b^{2}$

Rearrange to isolate $b^{2}$ :
$4 b^{2} - b^{2} = a^{2} - c^{2}$
$3 b^{2} = a^{2} - c^{2}$
$b^{2} = \frac{a ^{2} - c ^{2}}{3}$

$E x p r ess i o n f or cos A$

In $Δ A B C$ , using the Cosine Rule for $∠ A$ :
$cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$

$F in d in g t h e P r o d u c t cos A cos C$

Multiply the two expressions:
$cos A cos C = (\frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}) \times (\frac{2 b}{a})$
Simplify by canceling $2 b$ :
$cos A cos C = \frac{b ^{2} + c ^{2} - a ^{2}}{a c}$

$S u b s t i t u t in g b^{2}$

Substitute $b^{2} = \frac{a ^{2} - c ^{2}}{3}$ into the expression:
$cos A cos C = \frac{\frac{a ^{2} - c ^{2}}{3} + c ^{2} - a ^{2}}{a c}$

$S im pl i f y in g t h e N u m er a t or$

Take a common denominator of $3$ in the numerator:
$cos A cos C = \frac{\frac{a ^{2} - c ^{2} + 3 c ^{2} - 3 a ^{2}}{3}}{a c}$
Combine like terms:
$cos A cos C = \frac{\frac{2 c ^{2} - 2 a ^{2}}{3}}{a c}$

$F ina l R es u l t$

Bring the $3$ down to the main denominator:
$cos A cos C = \frac{2 c ^{2} - 2 a ^{2}}{3 a c}$
Factor out $2$ :
$cos A cos C = \frac{2 ( c ^{2} - a ^{2} )}{3 a c}$
The proof is complete.

00:00 / 00:00

The Sigma Insight: Properties of Triangles

Solution Diagram

Analyzing the Setup

Imagine you are standing in a vast, open field, and before you lies a triangle

A B C

. We are given that

D

is the midpoint of

B C

, and a mysterious condition:

A D ⊥ A C

This right angle at

A

is the key that unlocks the entire structure. Since

A D ⊥ A C

, we recognize that

Δ A C D

is a right-angled triangle.

In this triangle, the side

A C

(which we call

b

) is adjacent to angle

C

, and the side

C D

is the hypotenuse. Since

D

is the midpoint of

B C

, we know

C D = \frac{a}{2}

Thus, we can define

cos C

with beautiful simplicity:

cos C = \frac{A C}{C D} = \frac{b}{a /2} = \frac{2 b}{a}

The Power of the Cosine Rule

Now, let us look at the entire triangle

A B C

. We need a bridge between the sides

a, b, c

and the angles, which is provided by the Cosine Rule.

For any triangle, the Cosine Rule states:

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

We now have two different expressions for

cos C

. By equating them, we create an algebraic bridge:

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

With a quick cross-multiplication, we find

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

. This simplifies to

3 b^{2} = a^{2} - c^{2}

, or:

b^{2} = \frac{a ^{2} - c ^{2}}{3}

The Final Synthesis

Our goal is to find the product

cos A cos C

. We already have

cos C = \frac{2 b}{a}

. Applying the Cosine Rule for angle

A

Δ A B C

, we get:

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

Now, let us multiply these two expressions:

cos A cos C = (\frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}) \times (\frac{2 b}{a})

Watch closely as the

2 b

terms cancel out, leaving us with:

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

We substitute our earlier finding,

b^{2} = \frac{a ^{2} - c ^{2}}{3}

, into this expression:

cos A cos C = \frac{\frac{a ^{2} - c ^{2}}{3} + c ^{2} - a ^{2}}{a c}

By taking a common denominator of

3

in the numerator, we get:

\frac{a ^{2} - c ^{2} + 3 c ^{2} - 3 a ^{2}}{3 a c} = \frac{2 c ^{2} - 2 a ^{2}}{3 a c}

Finally, factoring out the

2

, we arrive at the elegant result:

$cos A cos C = \frac{2 ( c ^{2} - a ^{2} )}{3 a c}$

Common Traps & FAQs

Wait, why do we reject the negative root here? Or do we even need to?

Great observation! In this specific problem, we are dealing with lengths of sides of a triangle, which must be positive. When we derived

b^{2} = \frac{a ^{2} - c ^{2}}{3}

, we didn't need to take a square root because the expression for

cos A cos C

naturally uses

b^{2}

. We simply substitute the value of

b^{2}

directly, avoiding any ambiguity about signs.

Why can't we just use the Sine Rule instead of the Cosine Rule?

You absolutely could! The Sine Rule is a valid tool. However, the Cosine Rule is often more direct when you are working with side lengths and need to relate them to the cosine of an angle. Using the Sine Rule would likely introduce sines and potentially require more trigonometric identities to convert everything back to cosines. Stick with the Cosine Rule for this one—it's the most efficient path.

Is there a way to solve this using vectors?

Yes, absolutely! You could place vertex

A

at the origin

(0, 0)

and define the coordinates of

B

and

C

. Since

A D ⊥ A C

, you could use the dot product to represent the perpendicularity condition. While that is a powerful method, the geometric/trigonometric approach we used is often preferred in JEE for its elegance and speed. Both methods are valid, but mastering this trigonometric path will sharpen your intuition for similar problems.

Animated Solution for Mathematics - Trigonometry: $A B C$ is a triangle. $D$ is the middle point of $B C$ . If $A D$ is perpendicular to $A C$ , then prove that $cos A cos C = \frac{2 ( c ^{2} - a ^{2} )}{3 a c}$ .

Visualized Solution

$V i s u a l i z in g t h e T r ian g l e A B C$

$T r i g o n o m e t r y in Δ A C D$

$C os in e R u l e f or Δ A B C$

$E q u a t in g t h e E x p r ess i o n s$

$S o l v in g f or b^{2}$

$E x p r ess i o n f or cos A$

$F in d in g t h e P r o d u c t cos A cos C$

$S u b s t i t u t in g b^{2}$

$S im pl i f y in g t h e N u m er a t or$

$F ina l R es u l t$

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Power of the Cosine Rule

The Final Synthesis

Similar Questions

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

With usual notation, if in a triangle $A B C$ ; $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ , then prove that $\frac{c o s A}{7} = \frac{c o s B}{19} = \frac{c o s C}{25}$ .

$A B C$ is a triangle with $∠ B$ greater than $∠ C$ . $D$ and $E$ are points on $B C$ such that $A D$ is perpendicular to $B C$ and $A E$ is the bisector of angle $A$ . Complete the relation $∠ D A E = \frac{1}{2} [(\dots) - ∠ C]$

$A B C$ is a triangle with $A B = A C$ . $D$ is any point on the side $B C$ . $E$ and $F$ are points on the side $A B$ and $A C$ , respectively, such that $D E$ is parallel to $A C$ , and $D F$ is parallel to $A B$ . Prove that $D F + F A + A E + E D = A B + A C$ .

In a triangle $A B C$ , if $cos A + 2 cos B + cos C = 2$ and the lengths of the sides opposite to the angles $A$ and $C$ are 3 and 7 respectively, then $cos A - cos C$ is equal to

If in a triangle $A B C$ , $\frac{2 c o s A}{a} + \frac{c o s B}{b} + \frac{2 c o s C}{c} = \frac{a}{b c} + \frac{b}{c a}$ , then the value of the angle $A$ is .................. degrees.

$A B C$ is a triangle, $P$ is a point on $A B$ , and $Q$ is point on $A C$ such that $∠ A QP = ∠ A B C$ . Complete the relation $\frac{area of Δ A P Q}{area of Δ A B C} = \frac{( \dots )}{A C ^{2}}$

Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, then prove that cosAcosC=3ac2(c2−a2)​.

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solvingforb2

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitutingb2

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

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

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Power of the Cosine Rule

The Final Synthesis

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; } With usual notation, if in a triangle ABC; 11b+c​=12c+a​=13a+b​, then prove that 7cosA​=19cosB​=25cosC​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a triangle ABC, if cosA+2cosB+cosC=2 and the lengths of the sides opposite to the angles A and C are 3 and 7 respectively, then cosA−cosC is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If in a triangle ABC, a2cosA​+bcosB​+c2cosC​=bca​+cab​, then the value of the angle A is .................. degrees.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC is a triangle, P is a point on AB, and Q is point on AC such that ∠AQP=∠ABC. Complete the relation area of ΔABCarea of ΔAPQ​=AC2(…)​

Animated Solution for Mathematics - Trigonometry: $A B C$ is a triangle. $D$ is the middle point of $B C$ . If $A D$ is perpendicular to $A C$ , then prove that $cos A cos C = \frac{2 ( c ^{2} - a ^{2} )}{3 a c}$ .

$V i s u a l i z in g t h e T r ian g l e A B C$

$T r i g o n o m e t r y in Δ A C D$

$C os in e R u l e f or Δ A B C$

$E q u a t in g t h e E x p r ess i o n s$

$S o l v in g f or b^{2}$

$E x p r ess i o n f or cos A$

$F in d in g t h e P r o d u c t cos A cos C$

$S u b s t i t u t in g b^{2}$

$S im pl i f y in g t h e N u m er a t or$

$F ina l R es u l t$

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

With usual notation, if in a triangle $A B C$ ; $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ , then prove that $\frac{c o s A}{7} = \frac{c o s B}{19} = \frac{c o s C}{25}$ .

In a triangle $A B C$ , if $cos A + 2 cos B + cos C = 2$ and the lengths of the sides opposite to the angles $A$ and $C$ are 3 and 7 respectively, then $cos A - cos C$ is equal to

If in a triangle $A B C$ , $\frac{2 c o s A}{a} + \frac{c o s B}{b} + \frac{2 c o s C}{c} = \frac{a}{b c} + \frac{b}{c a}$ , then the value of the angle $A$ is .................. degrees.

$A B C$ is a triangle, $P$ is a point on $A B$ , and $Q$ is point on $A C$ such that $∠ A QP = ∠ A B C$ . Complete the relation $\frac{area of Δ A P Q}{area of Δ A B C} = \frac{( \dots )}{A C ^{2}}$