1.

In Δ ABC ∠B = ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX =70°, then ∠ACB =A. 35°B. 90°C. 70°D. 55°

Answer»

AX bisects ∠DAC

∠CAD = 2 x ∠DAC

∠CAD = 2 x 70°

= 140°

By exterior angle theorem,

∠CAD = ∠B + ∠C

140° = ∠C + ∠C (Therefore, ∠B = ∠C)

140° = 2∠C

∠C = 70°

Therefore,

∠C = ∠ACB = 70°



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