We are asked to calculate the size of angle $\theta$ in the isosceles triangle shown, with the two equal sides measuring 9.6 cm and the base measuring 6.8 cm.

Steps to solve for $\theta$:

1. Identify the known information:

   • We have an isosceles triangle where two sides are equal, each measuring 9.6 cm.
   • The base of the triangle is 6.8 cm.
   • The angle $\theta$ is the angle at the vertex between the two equal sides.

2. Apply trigonometry: Since the triangle is isosceles, we can split it into two right-angled triangles by drawing a height from the vertex (where $\theta$ is) perpendicular to the base. This height divides the base into two equal parts, each of $6.8 \div 2 = 3.4$ cm.

3. Use the cosine rule or trigonometric relationships: In one of the right triangles, we can apply the sine rule or cosine rule, or use basic trigonometry. Using the cosine rule would be efficient here:

   $\cos(\theta / 2) = \frac{\text{adjacent side (half the base)}}{\text{hypotenuse (one of the equal sides)}} = \frac{3.4}{9.6}$

   Solving this gives us:

   $\cos(\theta / 2) = \frac{3.4}{9.6}$

4. Calculate $\theta$: Now solve for $\theta / 2$, and then double it to find $\theta$.

Let me calculate this.The size of angle $\theta$ in the isosceles triangle is approximately $138.5^\circ$, rounded to one decimal place.

Would you like any more details or have any questions about this solution?

Here are some additional questions to explore:

1. How can you apply the sine rule to solve similar problems?
2. What is the relationship between the base angles in an isosceles triangle?
3. How would the solution change if the base or equal sides were different lengths?
4. Can this approach be applied to non-isosceles triangles?
5. What would be the area of this isosceles triangle?

Tip: When working with isosceles triangles, always consider symmetry to simplify the problem by splitting it into two right triangles.