Question

cho tam giác ABC có góc A = 105 độ, góc B = 45 độ; BC=4. Tính `AB`.

Answer 1

\(\widehat{C}=180^0-105^0-45^0=30^0\)

Xét ΔABC có \(\dfrac{BC}{sinA}=\dfrac{AB}{sinC}\)

=>\(\dfrac{AB}{sin30}=\dfrac{4}{sin105}\)

=>\(AB=2\sqrt{6}-2\sqrt{2}\left(cm\right)\)