Bài 4:
Ta có: $\widehat{A} = 60^\circ;\quad \widehat{B} = 45^\circ$
$\to \widehat{C} = 75^\circ$
Từ $A$ kẻ đường cao $AH$
$\to \begin{cases}AH = AC.\sin C = 2\sin75^\circ = \dfrac{1 + \sqrt3}{\sqrt2}\\HC = AC\cos75^\circ = 2\cos75^\circ = \dfrac{\sqrt3 - 1}{\sqrt2}\end{cases}$
$\to AH = HB = \dfrac{1+ \sqrt3}{\sqrt2}$
$\to AB = \dfrac{1+ \sqrt3}{\sqrt2}\cdot \sqrt2 = 1 + \sqrt3$
$\to BC = HB + HC = \dfrac{1 + \sqrt3}{\sqrt2} + \dfrac{\sqrt3 - 1}{\sqrt2} = \sqrt6$
Ta có:
$R = \dfrac{BC}{2\sin A} = \dfrac{\sqrt6}{2\sin60^\circ} = \sqrt2$
$S = \dfrac{AB.AC.BC}{4R} = \dfrac{(1+ \sqrt3).2.\sqrt6}{4\sqrt2} = \dfrac{3 + \sqrt3}{2}$
Bài 5:
$+)$ Áp dụng định lý $\cos$ ta được:
$BC^2 = AB^2 + AC^2 - 2AB.AC.\cos A$
$\to BC^2 = 5^2 + 7^2 - 2.5.7.\dfrac35$
$\to BC^2 = 32$
$\to BC = 4\sqrt2$
$+) \quad \cos A = \dfrac35 \to \sin A = \dfrac45$
$\to S = \dfrac12AB.AC.\sin A$
$\to S = \dfrac12.5.7.\dfrac45$
$\to S =14$
$+)\quad S = \dfrac12h_a.BC$
$\to h_a = \dfrac{2S}{BC}$
$\to h_a = \dfrac{2.14}{4\sqrt2}$
$\to h_a = \dfrac{7\sqrt2}{2}$
$+) \quad R = \dfrac{AB.AC.BC}{4S}$
$\to R = \dfrac{5.7.4\sqrt2}{4.14}$
$\to R = \dfrac{5\sqrt2}{2}$
