The diagram below shows two right-angled triangles, ABC and ACD - Junior Cycle Mathematics - Question 8 - 2017

From triangle $ABD$, we can use the Pythagorean theorem:

$|AC|^2 = |AB|^2 + |BC|^2$

Substituting the known values:

$12^2 = 10^2 + x^2$

Which simplifies to:

$144 = 100 + x^2$

So:

$x^2 = 144 - 100 = 44$

Therefore:

$x = \sqrt{44} \approx 6.633$

Rounding to two decimal places, we find that $x \approx 6.63$.

To find angle $T$, we can use the properties of the circle.

Given:

$\angle PQR = 22^{\circ}$

The relationship between angles subtended by the same arc states that:

$\angle PSR = \angle PQR$

Therefore:

$\angle PSR = 22^{\circ}$

Using the property that angles in the same segment are equal, we can express angle $T$:

$\angle T = 180^{\circ} - 2\times \angle PQR$

Calculating:

$\angle T = 180^{\circ} - 2\times 22^{\circ} = 180^{\circ} - 44^{\circ} = 136^{\circ}$

Thus, angle $T$ is 112 degrees.