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प्रश्न
In ΔABC, ∠B = ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX = 70°, then ∠ACB =
पर्याय
35°
90°
70°
55°
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उत्तर
In the given ΔABC, ∠B = ∠C . D is the ray extended from point A. AX bisects∠DAC and ∠DAX = 70°
Here, we need to find ∠ACB
As ray AX bisects ∠DAC
∠CAX = ∠DAX = 70°
Thus,
∠DAC = ∠DAX + ∠XAC
= 70° + 70°
= 140°
Now, according to the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”, we get,
∠DAC = ∠B + ∠C
140° = 2∠C
`∠C = (140°)/2`
= 70°
Thus, ∠ACB = 70°
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