Here is a pyramid with a square base ABCD - Edexcel - GCSE Maths - Question 13 - 2018 - Paper 3

Since ABCD is a square, the length of diagonal AC can be calculated using the Pythagorean theorem:

$AC = AB imes ext{sqrt}(2) = 5 imes ext{sqrt}(2)$

This gives us:

$AC ext{ length} = 5 ext{sqrt}(2)$.

The midpoint of AC is half of the length of AC:

Let M be the midpoint of AC. Hence, the distance AM (or CM) is:

$AM = \frac{AC}{2} = \frac{5\text{sqrt}(2)}{2}$.

Now using Pythagorean theorem for triangle ATM, we can calculate length TM (from T to the midpoint):

$TM = 12 ext{ m}$.

In triangle ATM, we can use tangent to find angle TAC:

$\tan(TAC) = \frac{TM}{AM} = \frac{12}{\frac{5\text{sqrt}(2)}{2}} = \frac{12 \times 2}{5\text{sqrt}(2)} = \frac{24}{5\text{sqrt}(2)}$.

Now, inverting the tangent:

$TAC = \tan^{-1}\left(\frac{24}{5\text{sqrt}(2)}\right)$.

Using a calculator, this gives approximately:

$\approx 74.1 \degree$. Hence, the answer is within the acceptable range of 73.5 to 74.1 degrees.