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

Q: Why can we assume $\sin C \le 1$?

Because $C$ is an angle in a triangle, $0 < C < 180^\circ$, so $\sin C$ is always between $0$ and $1$. This is a fundamental property of triangles that often acts as a hidden constraint in trigonometric equations.

Q: Why does $\cos(A - B) \ge 1$ force $A = B$?

Because the range of the cosine function is $[-1, 1]$. If $\cos(A - B) \ge 1$, it must be exactly $1$, which happens only when the argument is $0$. Thus, $A - B = 0$, implying $A = B$.

Q: Can we use the Sine Rule immediately?

The Sine Rule relates sides to angles, but we don't know the angles yet. We must solve the trigonometric constraint first to find the angles, and only then can we use the Sine Rule to find the ratio of the sides.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the vertex of a triangle, looking at a complex trigonometric equation: $\cos A \cos B + \sin A \sin B \sin C = 1$. In the world of JEE Advanced, complexity is often just a mask for a beautiful, hidden simplicity.

Our goal is to find the ratio $a : b : c = 1 : 1 : \sqrt{2}$. To do this, we must first untangle the relationship between the angles.

We start by isolating the term involving $\sin C$. By moving the cosine terms to the right, we get:

$$\sin A \sin B \sin C = 1 - \cos A \cos B$$

This is our first tactical maneuver, essentially clearing the battlefield to see the enemy more clearly.

### The Inequality Breakthrough

Now, we divide both sides by $\sin A \sin B$. Since $A$ and $B$ are angles of a triangle, their sines are strictly positive, so we do not need to worry about dividing by zero or flipping inequality signs.

We arrive at:

$$\sin C = \frac{1 - \cos A \cos B}{\sin A \sin B}$$

Here is the 'Aha!' moment. We know that for any angle $C$ in a triangle, $\sin C \le 1$. This is a universal truth.

By substituting our expression, we get:

$$\frac{1 - \cos A \cos B}{\sin A \sin B} \le 1$$

Multiplying by $\sin A \sin B$, we obtain $1 - \cos A \cos B \le \sin A \sin B$, which rearranges to:

$$1 \le \cos A \cos B + \sin A \sin B$$

### The Geometric Revelation

Look at that right side! It is the classic expansion for $\cos(A - B)$. So, our inequality becomes:

$$1 \le \cos(A - B)$$

But wait, we also know that $\cos(A - B) \le 1$. The only way both these conditions can be true is if $\cos(A - B) = 1$.

This forces $A - B = 0$, meaning $A = B$. Our triangle is **isosceles**.

Now, we substitute $B = A$ back into our original equation:

$$\cos^2 A + \sin^2 A \sin C = 1$$

Rearranging gives $\sin^2 A \sin C = 1 - \cos^2 A$. Since $1 - \cos^2 A = \sin^2 A$, we have:

$$\sin^2 A \sin C = \sin^2 A$$

Because $\sin A 
eq 0$, we can divide by $\sin^2 A$ to find $\sin C = 1$, which means $C = 90^\circ$.

### The Final Victory

We have discovered that our triangle is an isosceles right triangle with $C = 90^\circ$ and $A = B = 45^\circ$. Now, the Sine Rule is our final weapon:

$$\frac{a}{\sin 45^\circ} = \frac{b}{\sin 45^\circ} = \frac{c}{\sin 90^\circ}$$

Substituting the values, we get:

$$\frac{a}{\frac{1}{\sqrt{2}}} = \frac{b}{\frac{1}{\sqrt{2}}} = \frac{c}{1}$$

Multiplying the entire ratio by $\frac{1}{\sqrt{2}}$, we arrive at the elegant result:

**$a : b : c = 1 : 1 : \sqrt{2}$**

You have just conquered a complex trigonometric puzzle by breaking it down into simple, logical steps. Keep this mindset, and no problem will ever be too difficult for you.

Animated Solution for Mathematics - Trigonometry: If in a triangle $A B C$ , $cos A cos B + sin A sin B sin C = 1$ , Show that $a : b : c = 1 : 1 : 2$ .

Visualized Solution

Analyzing the Given Equation

Isolating $sin C$

Solving for $sin C$

Applying the Sine Bound

Rearranging the Inequality

Using Cosine Identity

The Equality Condition

Substituting $A = B$ Back

Simplifying the Expression

Finding Angle $C$

Calculating Angles $A$ and $B$

Applying the Sine Rule

Final Ratio Calculation

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Inequality Breakthrough

The Geometric Revelation

The Final Victory

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