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