设△ABC为锐角三角形,a,b,c分别为A,B,C所对的边且sin²A=sin(兀/3+B)sin(兀/3一B)+sin²B.(1)求角A.(2)若AB,AC=12,a=2√7,求b+C.
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![设△ABC为锐角三角形,a,b,c分别为A,B,C所对的边且sin²A=sin(兀/3+B)sin(兀/3一B)+sin²B.(1)求角A.(2)若AB,AC=12,a=2√7,求b+C.](/uploads/image/z/3849446-38-6.jpg?t=%E8%AE%BE%E2%96%B3ABC%E4%B8%BA%E9%94%90%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2%2Ca%2Cb%2Cc%E5%88%86%E5%88%AB%E4%B8%BAA%2CB%2CC%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%E4%B8%94sin%26%23178%3BA%3Dsin%28%E5%85%80%2F3%2BB%29sin%28%E5%85%80%2F3%E4%B8%80B%29%2Bsin%26%23178%3BB.%EF%BC%881%EF%BC%89%E6%B1%82%E8%A7%92A.%282%29%E8%8B%A5AB%2CAC%3D12%2Ca%3D2%E2%88%9A7%2C%E6%B1%82b%2BC.)
设△ABC为锐角三角形,a,b,c分别为A,B,C所对的边且sin²A=sin(兀/3+B)sin(兀/3一B)+sin²B.(1)求角A.(2)若AB,AC=12,a=2√7,求b+C.
设△ABC为锐角三角形,a,b,c分别为A,B,C所对的边且sin²A=sin(兀/3+B)sin(兀/3一B)+sin²B.(1)求角A.(2)若AB,AC=12,a=2√7,求b+C.
设△ABC为锐角三角形,a,b,c分别为A,B,C所对的边且sin²A=sin(兀/3+B)sin(兀/3一B)+sin²B.(1)求角A.(2)若AB,AC=12,a=2√7,求b+C.
为什没不发图?
(1)、sin²A=sin(兀/3+B)sin(兀/3一B)+sin²B
=(v3/2*cosB+1/2*sinB)(v3/2*cosB-1/2*sinB)+sin^2B
=3/4*cos^2-1/4*sin^2B+sin^2B
=3/4,
——》sinA=v3/2,
——》A=π/3,
(2)、题目应该是:AB*AC=12吧,若...
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(1)、sin²A=sin(兀/3+B)sin(兀/3一B)+sin²B
=(v3/2*cosB+1/2*sinB)(v3/2*cosB-1/2*sinB)+sin^2B
=3/4*cos^2-1/4*sin^2B+sin^2B
=3/4,
——》sinA=v3/2,
——》A=π/3,
(2)、题目应该是:AB*AC=12吧,若是,则:
由余弦定理:cosA=(b^2+c^2-a^2)/2bc
——》b^2+c^2=a^2+2bccosA=(2v7)^2+2*12*(1/2)=40,
——》(b+c)^2=b^2+c^2+2bc=40+12*2=64,
——》b+c=8。
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