In the triangle $ABC$, $AB = 8$ cm, $AC = 7$ cm, \angle ABC = 0.5$ radians and \angle ACB = x$ radians - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 2

Using the sine rule, we have:

$\frac{AB}{\sin(\angle ACB)} = \frac{AC}{\sin(\angle ABC)}$

Substituting the known values:

$\frac{8}{\sin x} = \frac{7}{\sin(0.5)}$

Calculating $\sin(0.5)$:

$\sin(0.5) \approx 0.4794$

Thus, substituting into the equation gives:

$\frac{8}{\sin x} = \frac{7}{0.4794}$

Cross multiplying:

$8 \cdot 0.4794 = 7 \cdot \sin x$

So,

$\sin x = \frac{8 \cdot 0.4794}{7} \approx 0.548$

Rounding to 3 decimal places:

$\sin x \approx 0.548$