M
ΔPQR is isosceles and therefore PQR  PRQ = 72°. ΔPSR is also isosceles therefore
RPS  RSP  36° (using the external angle theorem or by calculating that PRS  108°)
and x = 180° − 36° − 36° = 108°.
N
not to scale
P
xº
132
U
V
134
S
R
Q
T
As ST is parallel to UV, PRT = 132o(corresponding angles). So PRQ = 48o (angles on a
straight line). From the exterior angle of a triangle theorem, SQP = QPR + PRQ, so
x = 134 – 48 = 86
U
As QS = SR, SRQ = SQR = x°. So QST = 2x° (exterior angle theorem). Also TQS = 2x°
since QT = TS. As PT = QT, TPQ = TQP = 20°. Consider the interior angles of triangle
PQR: 20 + (20 + 2x + x) + x = 180. So 4x + 40 = 180, hence x = 35.
F
Angle PSR = 41° (opposite angles of a parallelogram are equal). Therefore x = 41 + 83 because the
exterior angle of a triangle is equal to the sum of the two interior opposite angles.
x°
S
y°
y°
z°
a°
b°
z°
The two angles marked y o are equal because they are in an isosceles triangle. For the same
reason, the angles z o are equal. Since an exterior angle of a triangle is the sum of the
two interior and opposite angles, it follows that a = 2y and b = 2z. Now a o + b o = 180 o since
they are the base angles of a parallelogram. So 2y + 2z = 180 giving y + z = 90. But, from the
angle sum of a triangle x + y + z = 180; hence x = 90.
Q
I
M
x
x
P
not to
scale
O
y y
R
N
As OQ is the reflection of OP in OM, QOM = POM; similarly, RON = PON.
Hence reflex QOR = 2 × MON = 260 o . Therefore QOR = 360 o − 260 o = 100 o .
S
Angles on a straight line add up to 180 o . The base angles of the triangle are 67 o and 71 o . The
angles of a triangle add to 180 o . Therefore x = 180 o − (67 o + 71 o )= 42 o .
H
The angles marked a°, c° and e° and may be considered to be the exterior angles of the triangle and
therefore have a total of 360°. As b° = a°, d° = c°, and f° = e° (all pairs of vertically opposite
angles), b + d + f = 360. So a + b + c + d + e + f = 720.
T
PMR  110° (adjacent angles on a straight line); ∠PRM = 50° (angle sum of a triangle) PRS 
130° (adjacent angles on a straight line).
(A shorter method uses the exterior angle theorem: PRS  MPR  PMR.)
A
Triangle PQR is equilateral; hence PRS = 60 o and therefore XSQ = 40 o + 60 o = 100 o
(exterior angle of a triangle is equal to the sum of the two interior opposite angles). Hence,
applying this result again, we deduce that SXT = (35 + 100) o = 135 o .