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Application of Sine and Cosine Theorems in Senior High School Mathematics
1, AC or BD can be answered, but BD is better.

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