Câu 14.
$A(1;-2);B(-2;3);C(0;4)$
`=>\vec{AB}=(-3;5)`
`=>AB=\sqrt{(-3)^2+5^2}=\sqrt{34}`
`\qquad \vec{AC}=(-1;6)`
`=>AC=\sqrt{(-1)^2+6^2}=\sqrt{37}`
`\qquad \vec{BC}=(2;1)`
`=>BC=\sqrt{2^2+1^2}=\sqrt{5}`
Áp dụng định lý Cosin, ta có:
`cosA={AB^2+AC^2-BC^2}/{2AB.AC}`
`={34+37-5}/{2.\sqrt{34}.\sqrt{37}}={33}/{\sqrt{1258}`
$\\$
Vì `0°<\hat{A}<180°`
`=>0<sin \hat{A}<1`
Ta có: `sin^2A+cos^2A=1`
`=>sin^2A=1-cos^2A=1-{33^2}/{1258}={169}/{1258}`
`=>sinA={13}/{\sqrt{1258}}` (vì $sinA>0)$
$\\$
`\qquad S_{∆ABC}=1/ 2 AB.AC.sinA`
`=1/ 2 . \sqrt{34}.\sqrt{37}.{13}/{\sqrt{1258}}={13}/2(đv dt)`
Đáp án $A$
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Câu 15.
$A(1;-1);B(3;-3);C(6;0)$
`=>\vec{AB}=(2;-2)`
`=>AB=\sqrt{2^2+(-2)^2}=\sqrt{8}=2\sqrt{2}`
`\qquad \vec{AC}=(5;1)`
`=>AC=\sqrt{5^2+1^2}=\sqrt{26}`
`\qquad \vec{BC}=(3;3)`
`=>BC=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}`
Áp dụng định lý Cosin, ta có:
`cosA={AB^2+AC^2-BC^2}/{2AB.AC}`
`={8+26-18}/{2.2\sqrt{2}.\sqrt{26}}=2/{\sqrt{13}`
$\\$
Vì `0°<\hat{A}<180°`
`=>0<sin \hat{A}<1`
Ta có: `sin^2A+cos^2A=1`
`=>sin^2A=1-cos^2A=1-{2^2}/{13}=9/{13}`
`=>sinA={3}/{\sqrt{13}}` (vì $sinA>0)$
$\\$
`\qquad S_{∆ABC}=1/ 2 AB.AC.sinA`
`=1/ 2 . 2\sqrt{2}.\sqrt{26}. 3/{\sqrt{13}}=6(đv dt`
Đáp án $B$
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Câu 16.
$A(1;1);B(2;4);C(10;-2)$
`=>\vec{AB}=(1;3);\vec{AC}=(9;-3)`
`=>\vec{AB}.\vec{AC}=1.9+3.(-3)=0`
`=>\vec{AB}`$\perp $`\vec{AC}`
`=>∆ABC` vuông tại $A$
`=>\hat{BAC}=90°`
Đáp án $A$
