In a triangle ABC, the median to the side BC is of length 1/√(11 - 6√3) and it divides the angle A i

Q: Why can't we just use the Apollonius Theorem to find the side lengths?

While Apollonius Theorem is a powerful tool for medians, it requires knowing the lengths of the other two sides ($AB$ and $AC$) or the angles between them in a way that often leads to a system of equations involving unknown side lengths. The Sine Rule is more direct here because it links the median length directly to the angles, which are already given or easily derived.

Q: How did you determine that $\angle C = 105^\circ - \theta$?

In any triangle, the sum of interior angles is $180^\circ$. We are given $\angle A = 30^\circ + 45^\circ = 75^\circ$. If we define $\angle B = \theta$, then $\angle C = 180^\circ - (\angle A + \angle B) = 180^\circ - (75^\circ + \theta) = 105^\circ - \theta$. This is a standard step in triangle geometry problems.

Q: Why does the complex square root term in the median length cancel out?

This is a classic feature of well-designed JEE problems. The problem is constructed so that the trigonometric result, specifically $\csc \theta$, yields a value that is the reciprocal of the irrational part of the median length. When you multiply them, the surds rationalize and cancel, leaving you with a clean, integer result.

Question

In a triangle $ABC$, the median to the side $BC$ is of length $\frac{1}{\sqrt{11 - 6\sqrt{3}}}$ and it divides the angle $A$ into angles $30^\circ$ and $45^\circ$. Find the length of the side $BC$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In triangle $ABC$, the median $AD$ divides the base $BC$ into two equal segments, $BD$ and $DC$. We are given the length of the median as:

$$AD = \frac{1}{\sqrt{11 - 6\sqrt{3}}}$$

The angles are provided as $\angle BAD = 30^\circ$ and $\angle CAD = 45^\circ$. Let $\angle B = 	heta$. Since the sum of angles in $\Delta ABC$ is $180^\circ$, we determine that $\angle C = 180^\circ - 75^\circ - 	heta = 105^\circ - 	heta$.

### Applying the Sine Rule

We apply the Sine Rule to the two smaller triangles, $\Delta ABD$ and $\Delta ACD$. For $\Delta ABD$:

$$\frac{BD}{\sin 30^\circ} = \frac{AD}{\sin 	heta} \Rightarrow BD = \frac{AD}{2 \sin 	heta}$$

For $\Delta ACD$:

$$\frac{DC}{\sin 45^\circ} = \frac{AD}{\sin(105^\circ - 	heta)} \Rightarrow DC = \frac{AD}{\sqrt{2} \sin(105^\circ - 	heta)}$$

### The Master Equation

Since $D$ is the midpoint of $BC$, we have $BD = DC$. Equating the two expressions and canceling $AD$, we obtain:

$$\frac{1}{2 \sin 	heta} = \frac{1}{\sqrt{2} \sin(105^\circ - 	heta)}$$

This simplifies to the trigonometric identity:

$$\sin(105^\circ - 	heta) = \sqrt{2} \sin 	heta$$

### Solving for the Base

Expanding the left side using the sine subtraction formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$, we get:

$$\sin 105^\circ \cos 	heta - \cos 105^\circ \sin 	heta = \sqrt{2} \sin 	heta$$

Dividing by $\sin 	heta$ and substituting the values $\sin 105^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$ and $\cos 105^\circ = \frac{\sqrt{2} - \sqrt{6}}{4}$, we find:

$$\cot 	heta = 3\sqrt{3} - 4$$

Using the identity $\csc^2 	heta = 1 + \cot^2 	heta$, we calculate $\csc 	heta = 2\sqrt{11 - 6\sqrt{3}}$. Substituting this back into the expression for $BD$:

$$BD = \frac{AD}{2} \cdot \csc 	heta = \frac{1}{2\sqrt{11 - 6\sqrt{3}}} \cdot 2\sqrt{11 - 6\sqrt{3}} = 1$$

Since $BC = BD + DC = 1 + 1$, the final length is **$BC = 2$ units**.

Animated Solution for Mathematics - Trigonometry: In a triangle $A B C$ , the median to the side $B C$ is of length $\frac{1}{11 - 6 3}$ and it divides the angle $A$ into angles $3 0^{\circ}$ and $4 5^{\circ}$ . Find the length of the side $B C$ .

Visualized Solution

Visualizing the Triangle and Median

Defining the Unknown Angles

Applying Sine Rule in $Δ A B D$

Applying Sine Rule in $Δ A C D$

Equating the Segments

Expanding and Solving for $cot θ$

Finding $csc θ$

Calculating $B D$

Final Answer: Length of $B C$

The Sigma Insight: Properties of Triangles

Analyzing the Setup

Applying the Sine Rule

The Master Equation

Solving for the Base

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Visualized Solution

Visualizing the Triangle and Median

Defining the Unknown Angles

Applying Sine Rule in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔABD

Applying Sine Rule in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔACD

Equating the Segments

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

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

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

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