Can someone solve this??

  see below

Step-by-step explanation:

There are a few relevant relations involved:

an inscribed angle is half the measure of the arc it interceptsan arc has the same measure as the central angle that intercepts itthe angle exterior to a circle where secants meet is half the difference of the intercepted arcs (near and far)the angle interior to a circle where secants meet is half the sum of the intercepted arcsthe angle where tangents meet is the supplement of the (near) arc interceptedan exterior angle of a triangle is equal to the sum of the remote interior anglesthe angle between a tangent and a radius is 90°the angle sum theorem

AB is a diameter, so arcs AB are 180°.

a) BC is the supplement to arc AC: 180° -140° = 40°

b) BG is the supplement to AG: 180° -64° -38° = 78°

c) ∠1 has the measure of BC: 40°

d) ∠2 is inscribed in a semicircle, so has measure 180°/2 = 90°

e) ∠3 is half the measure of arc AE: 64°/2 = 32°

f) ∠4 is half the sum of arcs AG and BC: ((64°+38°) +40°)/2 = 71°

g) ∠5 is half the difference of arcs AC and EG: (140° -38°)/2 = 51°

h) ∠6 is half the sum of arcs EAC and BG: ((140°+64°) +78°)/2 = 141°

i) ∠7 is the difference of exterior angle 4 and interior angle 1: 71° -40° = 31°

j) ∠8 is the measure of arc AC: 140°

k) ∠9 is the supplement to arc AC: 180° -140° = 40°

l) ∠10 is the complement of angle 7: 90° -31° = 59°

QR= 16

Step-by-step explanation:

HOPE THIS HELPS