∠1, ∠2, and ∠3 form a straight line. If m∠3 is six more than twice m∠1 and m∠2 is 27 less than m∠3, find m∠2.

Here is their argument. given the obtuse angle x, we make a quadrilateral abcd with ∠dab = x, and ∠abc = 90◦, and ad = bc. say the perpendicular bisector to dc meets the perpendicular bisector to ab at p. then pa = pb and pc = pd. so the triangles pad and pbc have equal sides and are congruent. thus ∠pad = ∠pbc. but pab is isosceles, hence ∠pab = ∠pba. subtracting, gives x = ∠pad−∠pab = ∠pbc −∠pba = 90◦. this is a preposterous conclusion – just where is the mistake in the "proof" and why does the argument break down there?

Find the midpoint m of z1=(8+5i) and z2=(12+3i)

An arc subtends a central angle measuring 7pi/4 radians what fraction of the circumference is this arc?