Solution
In $\triangle ABC$, we are given the following:
$$ \angle BAC = 50^\circ, \quad AB = 10\ \text{cm}, \quad BC = 9\ \text{cm}. $$
We are tasked with finding the two possible values of $\angle BCA = x^\circ$ to one decimal place.
Solution
We will use the sine rule, which states:
$$ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}. $$
Here:
$$ A = 50^\circ, \quad a = BC = 9, \quad b = AC\ (\text{unknown}), \quad c = AB = 10. $$
Let $C = x^\circ$ and
$$ B = 180^\circ - (A + C) = 130^\circ - x. $$
Step 1: Applying the Sine Rule
Using the sine rule:
$$ \frac{\sin A}{a} = \frac{\sin C}{c}. $$
Substitute the known values:
$$ \frac{\sin 50^\circ}{9} = \frac{\sin x}{10}. $$
Rearranging for $\sin x$:
$$ \sin x = 10 \cdot \frac{\sin 50^\circ}{9}. $$
Calculate $\sin 50^\circ$:
$$ \sin 50^\circ \approx 0.7660. $$
Thus:
$$ \sin x = 10 \cdot \frac{0.7660}{9} \approx 0.8511. $$
Step 2: Finding $x$
The sine function gives two possible angles between $0^\circ$ and $180^\circ$. Therefore:
$$ x = \arcsin(0.8511). $$
Calculate $\arcsin(0.8511)$:
$$ x \approx 58.3^\circ. $$
The second possible angle is:
$$ x = 180^\circ - 58.3^\circ = 121.7^\circ. $$
Final Answer
The two possible values of $x$ are:
$$ \boxed{ 58.3^\circ \text{ and } 121.7^\circ } $$
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