Because after BD, the triangle BCD is an isosceles triangle, and the angle CBD is easier to find =30 degrees.
Therefore, the angle ABD= 120-30=90 degrees.
Triangle ABD is a right triangle.
BCD area = (1/2) BC× CD× sin120 =1/2× 2× (√ 3/2) = √ 3.
From cosine theorem, BD=2√3 is found.
ABD area =( 1/2)ab×BD = 1/2×4×2√3 = 4√3。
Therefore, the required area =5√3.
2. According to sine theorem, calculate the angle ABD.
AD/sinABD=AB/sinBDA
sinABD=5√3/ 14
ABD is an acute angle, and COSABD =1114.
Therefore, cosbad =-cos (60+Abd) = sin60sinabd-cos60sababd =1/7.
According to cosine theorem, BD = √ (100+196-2×10×1/7) =16.
According to sine theorem, BC/sinBDC=BD/sin 135.
BC = 16/(√2/2)* sin 30 = 8√2≈ 1 1.3km