Question

若△ABC的内角满足：2∠A-∠B=60°,4∠A+∠C=300°则△ABC是什么三角形?急,求细解.

Answer 1

2∠A-∠B=60°

∠B = 2∠A-60°

4∠A+∠C=300°

4∠A+∠C+∠B =300°+2∠A-60°

2∠A+∠C+∠B =240°

又因为∠A+∠C+∠B =180°

所以 ∠A=60°

∠B = 2∠A-60° =60°

∠C=60°

所以

△ABC是等边三角形