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 find the value of $x$ using the sine rule, we use the formula:

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

Here, let:

• $a = 7$, corresponding to side $AC$
• $b = 8$, corresponding to side $AB$
• $A = 0.5$ radians (angle $ABC$)
• $B = x$ (angle $ACB$)

Substituting into the sine rule gives:

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

Rearranging this yields:

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

Calculating the value of $\sin(0.5)$, we find:

$\sin(0.5) \approx 0.478$

Now substituting in:

$\sin(x) = \frac{8 \cdot 0.478}{7} \approx 0.548$

Therefore, $x \approx \arcsin(0.548) \approx 0.582$ radians when calculated, which rounds to 3 decimal places as $0.582$.