AB and AC are two tangents drawn from an external point A to a circle of radius 6 cm. If ∠BAC = 60°, what is the value, in centimetres, of √(AB² + AC²)?

Introduction / Context:
This is a geometry problem involving tangents drawn from an external point to a circle. The tangents have equal length and form a given angle at the external point. You are asked to find the square root of the sum of the squares of their lengths, which essentially uses properties of isosceles triangles and the Pythagoras theorem, together with a key fact about tangent segments from a common external point.

Given Data / Assumptions:

• A circle has centre O and radius r = 6 cm.
• AB and AC are tangents drawn from an external point A to the circle.
• ∠BAC = 60°.
• We must find √(AB² + AC²).
• Since AB and AC are tangents from the same point, AB = AC.

Concept / Approach:
First, note that AB = AC because tangents from a common external point are equal in length. Thus we need √(AB² + AC²) = √(2AB²) = AB * √2. To find AB, we use the geometry of tangents and chords. The angle between two tangents from an external point is supplementary to the angle subtended by the chord joining the points of contact at the centre. We then use this central angle to find the chord length BC, and finally relate BC to AB in triangle ABC, which is isosceles.

Step-by-Step Solution:
Step 1: Let B and C be the points of contact on the circle, and O be the centre. Step 2: Tangent segments AB and AC are equal, so AB = AC. Step 3: The angle between tangents at A is ∠BAC = 60°. Step 4: The angle at the centre subtended by chord BC is ∠BOC = 180° - ∠BAC = 180° - 60° = 120°. Step 5: In triangle BOC, OB = OC = radius = 6 cm. It is an isosceles triangle. Step 6: Use the cosine rule to find BC: BC² = OB² + OC² - 2 * OB * OC * cos(∠BOC). Step 7: Substitute values: BC² = 6² + 6² - 2 * 6 * 6 * cos(120°). Step 8: cos(120°) = -1/2, so BC² = 36 + 36 - 72 * (-1/2) = 72 + 36 = 108, hence BC = 6√3. Step 9: Now consider triangle ABC. It is isosceles with AB = AC and included angle ∠BAC = 60° and side BC opposite this angle. Step 10: Apply cosine rule in triangle ABC: BC² = AB² + AC² - 2 * AB * AC * cos(60°). Step 11: Let AB = AC = x, then BC² = x² + x² - 2x² * (1 / 2) = 2x² - x² = x². Step 12: Thus BC = x, so x = BC = 6√3 and therefore AB = AC = 6√3. Step 13: Compute √(AB² + AC²) = √((6√3)² + (6√3)²) = √(108 + 108) = √216 = 6√6.

Verification / Alternative check:
As a check, we can compute AB² + AC² directly. With AB = AC = 6√3, we have AB² = AC² = 108, so the sum is 216. The square root of 216 can be simplified by factoring: 216 = 36 * 6, so √216 = √36 * √6 = 6√6. This matches our final result. Also, the relationship BC = AB in the isosceles triangle with vertex angle 60° is consistent with the law of cosines, further confirming the internal geometry.

Why Other Options Are Wrong:
The value 4√6 would correspond to a smaller tangent length and would not match the chord length we found. The values 9√3 and 8√3 do not simplify to √(AB² + AC²) when AB = AC = 6√3. The numeric option 12 is not compatible with the computed sum of squares. Only 6√6 is consistent with a careful application of geometry and the Pythagoras theorem.

Common Pitfalls:
Students often forget that angle between tangents at an external point is supplementary to the angle at the centre, leading to an incorrect central angle. Others confuse the lengths of the tangent segments and the radius, or misapply the cosine rule. Some also try to compute AB directly using right triangles with the centre without correctly identifying the angles. A clear diagram and consistent stepwise application of known formulas avoid these errors.

Final Answer:
Therefore, the value of √(AB² + AC²) is 6√6.