Find B and C of a triangle ABC, if b = 2 cm, c = 1 cm and `A=60^(@)`.

Options

A. `90^(@) and 45^(@)`
B. `90^(@) and 30^(@)`
C. `60^(@) and 30^(@)`
D. none of these

Answer

To find angles B and C of triangle ABC given that \( b = 2 \, \text{cm} \), \( c = 1 \, \text{cm} \), and \( A = 60^\circ \), we can use the cosine rule and the sine rule. Here’s a step-by-step solution:

### Step 1: Use the Cosine Rule to find side \( a \)

The cosine rule states:
\[ a^2 = b^2 + c^2 - 2bc \cdot \cos A \]

Substituting the known values:
- \( b = 2 \, \text{cm} \)
- \( c = 1 \, \text{cm} \)
- \( A = 60^\circ \)

We have:
\[ a^2 = 2^2 + 1^2 - 2 \cdot 2 \cdot 1 \cdot \cos(60^\circ) \]

Calculating:
\[ a^2 = 4 + 1 - 2 \cdot 2 \cdot 1 \cdot \frac{1}{2} \]
\[ a^2 = 4 + 1 - 2 \]
\[ a^2 = 3 \]
\[ a = \sqrt{3} \, \text{cm} \]

### Step 2: Use the Cosine Rule to find angle \( B \)

Now, we can find angle \( B \) using the cosine rule again:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

Substituting the known values:
- \( a = \sqrt{3} \)
- \( b = 2 \)
- \( c = 1 \)

We have:
\[ \cos B = \frac{(\sqrt{3})^2 + 1^2 - 2^2}{2 \cdot \sqrt{3} \cdot 1} \]
\[ \cos B = \frac{3 + 1 - 4}{2\sqrt{3}} \]
\[ \cos B = \frac{0}{2\sqrt{3}} = 0 \]

Since \( \cos B = 0 \), we find:
\[ B = 90^\circ \]

### Step 3: Use the Angle Sum Property to find angle \( C \)

We know that the sum of angles in a triangle is \( 180^\circ \):
\[ A + B + C = 180^\circ \]
Substituting the known values:
\[ 60^\circ + 90^\circ + C = 180^\circ \]
\[ C = 180^\circ - 150^\circ = 30^\circ \]

### Final Answer
Thus, the angles of triangle ABC are:
- \( B = 90^\circ \)
- \( C = 30^\circ \)

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