Geometry Practice Test 4

Given ∠P = 40°
AC = QC and CB = CR
∠P +∠Q+∠R = 180°
40° +∠1+∠2 = 180
∠1+∠2 = 140° ......(i)
In ΔAQC,

∠1+∠1+∠3 = 180°
2∠1+∠3 = 180° ......(ii)
In ΔBCR,
∠2 +∠2 +∠4 = 180°
2∠2 +∠4 = 180°
Adding (ii) and (iii),
2(∠1+∠2)+∠3+∠4 = 360° (from (i))
280 + ∠3 + ∠4 = 360°
∠3 + ∠ 4 = 80°
∠3+∠4 +∠5 = 180° (angles on straight line)
∠5 = 180° − 80° = 100°