[Urgent] Math problem-solving conditions for senior high school entrance examination: AB =100m ∠ PAB = 75 ∠ QAB = 45 ∠ PBA = 60 ∠ QBA = 90 for PQ and AP.

Because angle PAB=75 and angle PBA=60, that is, angle APB=45.

AP /sinQBA=AB/sinAPB, calculated by AB= 100m, AP=200 times the square root of 6.

PB= (cosine theorem, Pb 2 = pa 2+ab 2-2pa * abcopab) can be obtained in the same way.

Because the angle QAB = 45°, QBA = 90 and AQB = 45°, that is, the triangle QAB is an isosceles right triangle.

Angle pbq = 90°-60° = 30°.

So QB=AB= 100m multiplexing cosine theorem, PQ 2 = Pb 2+QB 2-2pb * QBCOSPBQ can get PQ=

It should be such an idea, because it involves the root sign, so there is no calculation result

Wish to be the first ~ ~