The question in the image appears to be about a triangle $\triangle ABC$, with the following details:

1. $BC = 18 \, \text{cm}$ (base of the triangle),
2. $\text{Area} = 90 \, \text{cm}^2$,
3. $\text{Height from A to BC} = h = 10 \, \text{cm}$,
4. The task is to find the size of $\angle ACB$.

Solution:

We use the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ Substituting the known values: $90 = \frac{1}{2} \times 18 \times h$ $90 = 9 \times h$ $h = \frac{90}{9} = 10 \, \text{cm}$

Given that $h = 10 \, \text{cm}$, this confirms the triangle is valid.

To find $\angle ACB$: We use trigonometry, particularly if this is a right triangle:

1. If $ACB$ is $90^\circ$, the perpendicular height matches the triangle’s constraints.

Thus: $\angle ACB = 90^\circ$

If further clarification is needed on trigonometry steps, let me know.

Would you like additional clarification or details?

Questions:

1. What are the implications if $\triangle ABC$ is not a right triangle?
2. Can we verify this triangle’s dimensions using other trigonometric rules?
3. Would you like to explore sine, cosine, or tangent relationships here?
4. Do you need visualization of this triangle with dimensions?
5. Should I explain another mathematical approach to verify $\angle ACB$?

Tip:

Always cross-check the area and height relationships to verify the triangle's dimensions.