<span>You have mentioned that two of the angles of ΔKSD are
∠K=16° and ∠D=34°. So that would mean that ∠S=130°,
because∠K+∠D+∠S=180°. That's because all the angles
of any triangle must total 180°.
But you also say that ∠T=51°. But since no angle of
ΔKSD can equal 51°, neither can any angle of ΔWAT
equal to 51°. Therefore ∠T=51° is impossible!</span>
Answer:
<o=180-132=48
so,<s=24(The angle at the circumference is half of its corresponding angle at center)
