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 9 - 2005 - Paper 2

To apply the sine rule, we use the formula:

$\frac{a}{\sin A} = \frac{b}{\sin B}$

For triangle $ABC$, we have:

• $AB = 8$ cm (opposite $\angle ACB$)
• $AC = 7$ cm (opposite $\angle ABC$)

We can rearrange this to find $\sin x$:

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

Now we calculate $\sin(0.5)$:

$\sin(0.5) \approx 0.4794$

So substituting this value into the equation:

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

This simplifies to:

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

Finally, we find $x$ using the inverse sine function:

$x = \arcsin(0.548) \approx 0.582 \text{ radians (to 3 decimal places)}$