(a) 1 If is a right-angled at $$ \angle B $$ and $$ b=10 \mathrm{~cm} $$., then $$ \frac{a}{\sin A}+\frac{c}{\sin C}= $$

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Let's analyze the given problem step by step.

**Given:**
- Triangle $$ ABC $$ is right-angled at $$ \angle B $$.
- Side $$ b = 10 $$ cm (where $$ b $$ is opposite $$ \angle B $$).

**To Find:**
$$ \frac{a}{\sin A} + \frac{c}{\sin C} $$

**Solution:**

1. **Understanding the Triangle:**
   - Since $$ \angle B = 90^\circ $$, side $$ b $$ is the hypotenuse.
   - Angles $$ A $$ and $$ C $$ are complementary, i.e., $$ \angle A + \angle C = 90^\circ $$.
   - Sides $$ a $$ and $$ c $$ are the legs of the triangle.

2. **Applying the Sine Rule:**
   In any triangle, the sine rule states:
   $$
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
   $$
   
   Given that $$ \angle B = 90^\circ $$, we have:
   $$
   \sin B = \sin 90^\circ = 1
   $$
   
   Therefore:
   $$
   \frac{b}{\sin B} = \frac{10}{1} = 10 \text{ cm}
   $$
   
   This means:
   $$
   \frac{a}{\sin A} = 10 \text{ cm} \quad \text{and} \quad \frac{c}{\sin C} = 10 \text{ cm}
   $$

3. **Calculating the Sum:**
   $$
   \frac{a}{\sin A} + \frac{c}{\sin C} = 10 \text{ cm} + 10 \text{ cm} = 20 \text{ cm}
   $$

**Answer:**
$$
\frac{a}{\sin A} + \frac{c}{\sin C} = 20 \text{ cm}
$$
$$ \frac{a}{\sin A} + \frac{c}{\sin C} = 20 \text{ cm} $$

(a) 1 If is a right-angled at \( \angle B \) and \( b=10 \mathrm{~cm} \)., then \( \frac{a}{\sin A}+\frac{c}{\sin C}= \)

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