a)Xét tam giác vuông \(ABH\):

• Vuông tại \(H\),
• Góc \(\hat{B A H} = 28^{\circ}\),
• \(A H = 4\).

Ta có:

\(tan ⁡ 28^{\circ} = \frac{H B}{A H} \Rightarrow H B = A H \cdot tan ⁡ 28^{\circ}\) \(H B = 4 \cdot tan ⁡ 28^{\circ} \approx 4 \cdot 0 , 5317 = 2 , 13 \textrm{ } \text{cm}\)

Xét tam giác vuông \(A H C\):

• Vuông tại \(H\),
• Góc \(\hat{H C A} = 41^{\circ}\),
• \(A H = 4\).

Ta có:

\(tan ⁡ 41^{\circ} = \frac{A H}{H C} \Rightarrow H C = \frac{A H}{tan ⁡ 41^{\circ}}\) \(H C = \frac{4}{0 , 8693} \approx 4 , 60 \textrm{ } \text{cm}\)

b) Trong tam giác vuông \(A H C\):

\(cos ⁡ 41^{\circ} = \frac{H C}{A C} \Rightarrow A C = \frac{H C}{cos ⁡ 41^{\circ}}\) \(A C = \frac{4 , 60}{0 , 7547} \approx 6 , 10 \textrm{ } \text{cm}\)

Vậy a) \(H B \approx 2 , 13 \textrm{ } \text{cm} , \textrm{ }\textrm{ } H C \approx 4 , 60 \textrm{ } \text{cm}\)
b) \(A H = 4 \textrm{ } \text{cm} , \textrm{ }\textrm{ } A C \approx 6 , 10 \textrm{ } \text{cm}\)

\(\) \(\)

Ta có : A+B+C=180 nên A =180 - ( B+C )= 180- 105=75

Áp dụng định lý Sin : a/sin A = b/ sin B = c / sin c

( BC = a , AC = b , AB = c)

AC = BC/ sin A = AC/sin B = 4,2/sin 75=AC/sin 65 = 3,94cm

AB=AC/sin B = AB/sin c = 3,94/sin 65 = AB / sin 40 = 2,79 cm

Vậy góc A = 75 , AC=3,94cm ,AB=2,79cm

Ta có : A + B + C = 180 suy ra A = 180 - B - C = 180 - 65 - 45 =70

AC = AB/ sin C = AC/sin B = 2,8/sin 45 = AC/sin 65 = 3,59 cm

BC=BC/sinA=AC/sin 65=BC/sin 70 = 3,59/sin 65=3,72 cm

Vậy góc A = 70, AC=3,59cm , BC = 3,72cm