P在△ABC边BC上,PC=2PB,已知∠ABC=45°,∠APC=60°,求∠ACB
∵ ∠ABC=45°, ∴ AD = BD = 1 + X,( 1 + x )/x = tan∠APC = tan60° = √3;
∴ x = ( √3 + 1 );
tan∠C = ( 1 + x )/( 2 - x ) = 2 + √3;∴ ∠ACB = 75° 。
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∵ ∠ABC=45°, ∴ AD = BD = 1 + X,( 1 + x )/x = tan∠APC = tan60° = √3;
∴ x = ( √3 + 1 );
tan∠C = ( 1 + x )/( 2 - x ) = 2 + √3;∴ ∠ACB = 75° 。