Problem 10.5 Let P and Q be points on the side AB of the triangle ABC (with P between A and Q) such that ZACP = ZQCB, and let AD be the angle bisector of ZBAC. Line AD meets lines CP and CQ at M andN respectively. Given that PN = CD and 3ZBAC = 2ZBCA, prove that triangles CQD and QNB have ZPCQ the same area.

Q: 3. Work out the angles marked x, y and z. D. 50° 52° 72° A B.

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A: Here the given angle is -150°. θ=-150°=-5π6=-5π6+2π=-5π+12π6=7π6

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A: We can find the area of the triangle as below

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A: our objective is to prove.

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A: a)  area=27.88 b) angle between vectors = 0 degree