đặt : A B = c AB=c A B = c ; A C = b AC=b A C = b ; B C = a BC=a B C = a và A H = h a AH=h_a A H = h a
ta có : A ^ = 180 − ( B ^ + C ^ ) = 180 − ( 30 + 40 ) = 110 \widehat{A}=180-\left(\widehat{B}+\widehat{C}\right)=180-\left(30+40\right)=110 A = 1 8 0 − ( B + C ) = 1 8 0 − ( 3 0 + 4 0 ) = 1 1 0
ta lại có : b s i n B = a s i n A ⇔ 12 s i n 30 = a s i n 110 ⇔ a = 12 s i n 30 . s i n 110 \dfrac{b}{sinB}=\dfrac{a}{sinA}\Leftrightarrow\dfrac{12}{sin30}=\dfrac{a}{sin110}\Leftrightarrow a=\dfrac{12}{sin30}.sin110 s i n B b = s i n A a ⇔ s i n 3 0 1 2 = s i n 1 1 0 a ⇔ a = s i n 3 0 1 2 . s i n 1 1 0
⇔ a ≃ 22 , 55 \Leftrightarrow a\simeq22,55 ⇔ a ≃ 2 2 , 5 5
ta có : S Δ A B C = 1 2 a . b . s i n C = 1 2 . ( 22 , 55 ) . 12. s i n 40 ≃ 86 , 97 S_{\Delta ABC}=\dfrac{1}{2}a.b.sinC=\dfrac{1}{2}.\left(22,55\right).12.sin40\simeq86,97 S Δ A B C = 2 1 a . b . s i n C = 2 1 . ( 2 2 , 5 5 ) . 1 2 . s i n 4 0 ≃ 8 6 , 9 7
ta có : S Δ A B C = 1 2 a . h a ⇔ 86 , 97 = 1 2 . ( 22 , 55 ) . h a ⇔ h a = 86 , 97. 2 22 , 55 ≃ 7 , 71 S_{\Delta ABC}=\dfrac{1}{2}a.h_a\Leftrightarrow86,97=\dfrac{1}{2}.\left(22,55\right).h_a\Leftrightarrow h_a=86,97.\dfrac{2}{22,55}\simeq7,71 S Δ A B C = 2 1 a . h a ⇔ 8 6 , 9 7 = 2 1 . ( 2 2 , 5 5 ) . h a ⇔ h a = 8 6 , 9 7 . 2 2 , 5 5 2 ≃ 7 , 7 1
vậy A H = 7 , 71 AH=7,71 A H = 7 , 7 1 và B C = 22 , 55 BC=22,55 B C = 2 2 , 5 5