题目内容
在△ABC中,a=3,b=2
,∠B=2∠A,
(1)求cos A的值;
(2)求c的值.
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(1)求cos A的值;
(2)求c的值.
(1)
(2)5
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(1)因为a=3,b=2
,∠B=2∠A,
所以在△ABC中,由正弦定理得
=
.
所以
=
.故cos A=
.
(2)由(1)知cos A=
,所以sin A=
=
.
又因为∠B=2∠A,所以cos B=2cos2A-1=
.
所以sin B=
=
.
在△ABC中,sin C=sin(A+B)=sin Acos B+
cos Asin B=
.
所以c=
=5.
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所以在△ABC中,由正弦定理得
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所以
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(2)由(1)知cos A=
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又因为∠B=2∠A,所以cos B=2cos2A-1=
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所以sin B=
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在△ABC中,sin C=sin(A+B)=sin Acos B+
cos Asin B=
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所以c=
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