In △ ABC, if B = 60 degrees, ab = 2, AC = 2 √ 3, then the area of △ ABC is ?

In △ ABC, if B = 60 degrees, ab = 2, AC = 2 √ 3, then the area of △ ABC is ?

AC/AB=√3/3=tanB
So the triangle ABC is a right triangle
Area = AC * AB / 2 = 2 √ 3

In △ ABC, ab = √ 3, AC = 1, B = 30 & # 186;, then the area of △ ABC is equal to

A=90º 1x√3x1/2=√3/2
The root of two is three

In △ ABC, if AB = 2 √ 3, AC = 6, ∠ a = 60 °, the area of △ ABC is
Such as the title

From the meaning of the title
S=(AB*AC*sin∠A)/2=9

In △ ABC, ∠ B = 120 °, AC = 7, ab = 5, then the area of △ ABC is______ ．

From the cosine theorem, we can know that CoSb = 25 + BC2 − 492 · BC · 5 = - 12, and the area of BC = - 8 or 3 (rounding off) ∧ ABC is 12 · ab · BC · SINB = 12 × 5 × 3 × 32 = 1534, so the answer is: 1534