cho tam giác ABC có cạnh ab=a,AC=a căn 3,BC=a căn 7.tính góc BAC

`cos \hat{BAC} = \frac{AB^{2} + AC^{2} - BC^{2}}{2.AB.AC}`

                  `= \frac{a^{2} + 3a^{2} - 7a^{2}}{2.a.a\sqrt{3}}`

                  `= \frac{- 3a^{2}}{2\sqrt{3}a^{2}}`

                  `= - \frac{\sqrt{3}}{2}`

`=> \hat{BAC} = 150^{0}`

` Vậy ` _________________________________________

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