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九年级下册数学,求解救. (1)COS60°-COS45°+tan45°, (2)cos²60°+sin²45°,
根据条件求锐角。
(1)sinA=0.753,求角A
(2)cosB=0.0832,求∠B
(3)tanC=45.8求∠C
人气:444 ℃ 时间:2020-04-20 07:34:20
解答
(1)COS60°-COS45°+tan45°,
=1/2-√2/2+1
=(3-√2)/2
(2)cos²60°+sin²45°,
=(1/2)²+(√2/2)²
=1/4+2/4
=3/4cos30°-sin45°/sin60°-cos45°,(cos30°-sin45°)/(sin60°-cos45°)=[cos(90°-30°)-sin(90°-45°)]/(sin60°-cos45°)=(sin60°-cos45°)/(sin60°-cos45°)=1(1)sinA=0.753,求角A(2)cosB=0.0832,求∠B(3)tanC=45.8求∠C(1)sinA=0.753,∠A=48.85°或∠A=131.15°(2)cosB=0.0832,∠B=85.23°(3)tanC=45.8∠C=88.75°不谢
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