In `DeltaABC`, right angled at B, AB = 5 cm and `angleACB==30^(@)`. Find BC and AC.

Answer

To solve the problem step by step, we will use the properties of right-angled triangles and trigonometric ratios.

### Step 1: Identify the triangle and given values
We have a right-angled triangle ABC, where:
- Angle B is 90 degrees (right angle).
- Side AB (the side opposite angle C) = 5 cm.
- Angle ACB = 30 degrees.

### Step 2: Use the tangent function to find BC
In triangle ABC, we can use the tangent function, which is defined as:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
For angle ACB (30 degrees), the opposite side is AB (5 cm) and the adjacent side is BC. Therefore:
\[ \tan(30^\circ) = \frac{AB}{BC} \]
Substituting the known values:
\[ \tan(30^\circ) = \frac{5}{BC} \]
We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\). So we can write:
\[ \frac{1}{\sqrt{3}} = \frac{5}{BC} \]
Cross-multiplying gives:
\[ BC = 5\sqrt{3} \text{ cm} \]

### Step 3: Use the sine function to find AC
Now, we will use the sine function to find the hypotenuse AC. The sine function is defined as:
\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \]
For angle ACB (30 degrees), the opposite side is AB (5 cm) and the hypotenuse is AC. Therefore:
\[ \sin(30^\circ) = \frac{AB}{AC} \]
Substituting the known values:
\[ \sin(30^\circ) = \frac{5}{AC} \]
We know that \(\sin(30^\circ) = \frac{1}{2}\). So we can write:
\[ \frac{1}{2} = \frac{5}{AC} \]
Cross-multiplying gives:
\[ AC = 10 \text{ cm} \]

### Final Results
- BC = \(5\sqrt{3} \text{ cm}\)
- AC = \(10 \text{ cm}\)