In `Delta PQR,PQ = 6 sqrt 3 cm, PR = 12cm and QR = 6 cm` then `angle Q` is

Options

A. `45^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Answer

To find the angle Q in triangle PQR where PQ = 6√3 cm, PR = 12 cm, and QR = 6 cm, we can use the cosine rule. The cosine rule states that for any triangle with sides a, b, and c opposite to angles A, B, and C respectively:

\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

In our case, we need to find angle Q, which is opposite side PR. Let's assign the sides as follows:
- PQ = a = 6√3 cm
- QR = b = 6 cm
- PR = c = 12 cm

Now, we can apply the cosine rule to find cos Q:

### Step 1: Write the cosine rule formula for angle Q
\[ \cos Q = \frac{PQ^2 + QR^2 - PR^2}{2 \cdot PQ \cdot QR} \]

### Step 2: Substitute the values into the formula
\[ \cos Q = \frac{(6\sqrt{3})^2 + (6)^2 - (12)^2}{2 \cdot (6\sqrt{3}) \cdot (6)} \]

### Step 3: Calculate the squares of the sides
\[ (6\sqrt{3})^2 = 36 \cdot 3 = 108 \]
\[ (6)^2 = 36 \]
\[ (12)^2 = 144 \]

### Step 4: Substitute the squared values back into the formula
\[ \cos Q = \frac{108 + 36 - 144}{2 \cdot (6\sqrt{3}) \cdot (6)} \]

### Step 5: Simplify the numerator
\[ \cos Q = \frac{0}{2 \cdot (6\sqrt{3}) \cdot (6)} \]

### Step 6: Calculate the denominator
\[ 2 \cdot (6\sqrt{3}) \cdot (6) = 72\sqrt{3} \]

### Step 7: Final calculation of cos Q
\[ \cos Q = \frac{0}{72\sqrt{3}} = 0 \]

### Step 8: Determine angle Q
Since \(\cos Q = 0\), we know that:
\[ Q = 90^\circ \]

### Final Answer:
Thus, angle Q is \(90^\circ\).

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