Discussion

Question:

The internal bisector of an angle A in a triangle ABC meets the side BC at point D. AB = 4, AC = 3 and ∠A = 60°. Then what is the length of the bisector AD?

Explanation:

Area of ∆ADC = $\frac{1}{2}$ AD × 3 × sin 30°

Area of ∆ADC = $\frac{1}{2}$ AD × 4 × sin 30°

Area of ∆ADC = $\frac{1}{2}$ × 3 × 4 × sin 60°

$\frac{1}{2}$ AD × 3 × sin 30° + $\frac{1}{2}$ × AD × 4 × sin 30° = $\frac{1}{2}$ × 3 × 4 × sin 60°

AD $[3 \times \frac{1}{2} + 4 \times \frac{1}{2}] = 12 \times \frac{\sqrt{3}}{2}$

AD $\frac{7}{2} = \frac{12 \sqrt{3}}{2}$

∴ AD = $\frac{12 \sqrt{3}}{7}$

Hence, option (a).

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