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In a /\ABC if a^(2)-b^(2)-c^(2)+(x-2)bc=...

In a `/_\ABC` if `a^(2)-b^(2)-c^(2)+(x-2)bc=0` then `x` be equal to

A

`1/2`

B

`3/2`

C

`2`

D

`4`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the value of \( x \) in the equation given for triangle \( \triangle ABC \): \[ a^2 - b^2 - c^2 + (x - 2)bc = 0 \] ### Step-by-Step Solution: 1. **Rearranging the Equation**: Start by isolating \( x \) in the equation: \[ a^2 - b^2 - c^2 + (x - 2)bc = 0 \] Rearranging gives: \[ (x - 2)bc = b^2 + c^2 - a^2 \] 2. **Dividing by \( bc \)**: Next, divide both sides by \( bc \) (assuming \( bc \neq 0 \)): \[ x - 2 = \frac{b^2 + c^2 - a^2}{bc} \] 3. **Adding 2 to Both Sides**: Now, add 2 to both sides to solve for \( x \): \[ x = \frac{b^2 + c^2 - a^2}{bc} + 2 \] 4. **Using the Cosine Rule**: According to the cosine rule in triangle \( ABC \): \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \] Therefore, we can express \( \frac{b^2 + c^2 - a^2}{bc} \) in terms of \( \cos A \): \[ \frac{b^2 + c^2 - a^2}{bc} = 2 \cos A \] 5. **Substituting Back**: Substitute this back into the equation for \( x \): \[ x = 2 \cos A + 2 \] 6. **Finding the Range of \( x \)**: Since \( \cos A \) can take values from -1 to 1 (as \( A \) varies from \( 0 \) to \( \pi \)): - When \( \cos A = -1 \): \[ x = 2(-1) + 2 = 0 \] - When \( \cos A = 1 \): \[ x = 2(1) + 2 = 4 \] Thus, the value of \( x \) lies in the range: \[ 0 < x < 4 \] ### Conclusion: The possible values for \( x \) are in the interval \( (0, 4) \).
To solve the problem, we need to find the value of \( x \) in the equation given for triangle \( \triangle ABC \): \[ a^2 - b^2 - c^2 + (x - 2)bc = 0 \] ### Step-by-Step Solution: ...
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