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In a triagle Delta ABC, the angle bisect...

In a triagle `Delta` ABC, the angle bisectors of `angle ABC` and `angle ACB` meet at point D and angle `angle BDC` is `135^@` . Find the value of `angle BAC`.

A

`45^@`

B

`75^@`

C

`60^@`

D

`90^@`

Text Solution

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The correct Answer is:
To solve the problem step by step, we will analyze the triangle and the given information about the angle bisectors and angles. ### Step 1: Understand the Triangle and the Angles We have triangle \( ABC \) where: - \( D \) is the intersection of the angle bisectors of \( \angle ABC \) and \( \angle ACB \). - We know that \( \angle BDC = 135^\circ \). ### Step 2: Set Up the Angles Let \( \angle BAC = A \). Since \( D \) is the intersection of the angle bisectors, we can express the angles at point \( D \): - \( \angle ABD = \frac{B}{2} \) (where \( B = \angle ABC \)) - \( \angle ACD = \frac{C}{2} \) (where \( C = \angle ACB \)) ### Step 3: Use the Angle Sum Property In triangle \( BDC \): \[ \angle BDC + \angle ABD + \angle ACD = 180^\circ \] Substituting the known values: \[ 135^\circ + \frac{B}{2} + \frac{C}{2} = 180^\circ \] ### Step 4: Simplify the Equation Rearranging the equation gives: \[ \frac{B}{2} + \frac{C}{2} = 180^\circ - 135^\circ \] \[ \frac{B}{2} + \frac{C}{2} = 45^\circ \] Multiplying through by 2: \[ B + C = 90^\circ \] ### Step 5: Relate Angles in Triangle ABC Since the sum of angles in triangle \( ABC \) is \( 180^\circ \): \[ A + B + C = 180^\circ \] Substituting \( B + C = 90^\circ \) into this equation: \[ A + 90^\circ = 180^\circ \] Solving for \( A \): \[ A = 180^\circ - 90^\circ = 90^\circ \] ### Final Answer Thus, the value of \( \angle BAC \) is \( 90^\circ \).
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