In a Triangle ABC, AB = 8 cm, AC = 6 cm and ∠BAC = 60°. Find the Length of the Median From Vertex A to Side BC

A standard triangle geometry problem, not tied to a specific exam paper.

Problem

In a triangle $ABC$, $AB=8$ cm, $AC=6$ cm, and $\angle BAC=60^\circ$. Find the length of the median from vertex $A$ to side $BC$.

Key insight. The median formula needs the length of the opposite side, not the two sides adjacent to the vertex the median starts from — so the given angle has to be used first, via the cosine rule, purely to find $BC$, before the median formula can even be applied.

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Approach

The median from $A$ to $BC$ depends on all three sides of the triangle, but only two sides ($AB$ and $AC$) and the included angle are given. First use the cosine rule on the angle at $A$ to find the missing side $BC$. Then apply the standard median-length formula, which expresses the median from a vertex in terms of the two adjacent sides and the length of the opposite side.

Solution

Step 1 — Label the sides using standard notation

Let $c=AB=8$, $b=AC=6$, and $a=BC$ (the side opposite $A$, currently unknown). The included angle is $\angle A=60^\circ$.

Step 2 — Find BC using the cosine rule

$$a^2 = b^2+c^2-2bc\cos A$$

$$a^2 = 6^2+8^2-2(6)(8)\cos60^\circ = 36+64-96\left(\frac12\right) = 100-48 = 52$$

So $BC^2=52$ — this value is needed, not $BC$ itself, so there’s no need to simplify $\sqrt{52}$ separately.

Step 3 — Apply the median length formula

The length of the median $m_a$ from vertex $A$ to side $BC=a$ is given by:

$$m_a^2 = \frac{2b^2+2c^2-a^2}4$$

This formula comes from doubling the triangle across the midpoint of $BC$ (Apollonius’s theorem) and expresses the median purely in terms of the three side lengths.

Step 4 — Substitute the known values

$$m_a^2 = \frac{2(6^2)+2(8^2)-52}4 = \frac{2(36)+2(64)-52}4 = \frac{72+128-52}4 = \frac{148}4 = 37$$

Step 5 — Take the square root

$$m_a = \sqrt{37}$$

Answer

Common mistakes

• Applying the median formula with the wrong side as “a”. The median formula $m_a^2=\dfrac{2b^2+2c^2-a^2}4$ requires $a$ to be the side opposite the vertex the median is drawn from — using $AB$ or $AC$ in that slot instead of the newly-found $BC$ gives a completely different (and wrong) result.
• Simplifying $\sqrt{52}$ unnecessarily. Since only $a^2=52$ is needed for the median formula, not $a$ itself, taking a square root here and squaring it again later just adds an unnecessary — and error-prone — extra step.
• Using the wrong sign or forgetting the factor of 2 in the cosine rule. $a^2=b^2+c^2-2bc\cos A$ has a minus sign in front of the cosine term; flipping it to a plus (as if using the sine rule’s structure) gives $a^2=148$ instead of $52$, and the whole answer changes.

Practise next

• In a triangle $ABC$, $AB=10$ cm, $AC=8$ cm, and $\angle BAC=60^\circ$. Find the length of the median from vertex $A$ to side $BC$.