<A and <B are complimentary angles of right triangle ABC cis A = 0.83 and cis B = 0.55 what is sin A + sin B A. 0.28
B. 1
C. 0.38
D. 1.38

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?

Consider the diagram. which line segment has the same measure as st? rx tx sr xs

The center of a circle represent by the equation (x+9)^2+(y-6)^2=10^2 (-9,6), (-6,9), (6,-9) ,(9,-6)