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In Figure, lines AB and CD intersect at O. If
∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find ∠ BOE and reflex ∠ COE
AB is a straight line and the rays OC and OE stand on it.
Reflex
CD is a straight line. Rays OE and OD stand on it.
In Figure, lines XY and MN intersect at O. If
∠ POY = 90° and a : b = 2 : 3, find c.
Let the common ratio between a and b be x.
∴ a = 2x, and b = 3x
XY is a straight line, rays OM and OP stand on it.
∴ ∠XOM + ∠MOP + ∠POY = 180º
b + a + ∠POY = 180º
3x + 2x + 90º = 180º
5x = 90º
x = 18º
a = 2x = 2 × 18 = 36º
b = 3x= 3 ×18 = 54º
MN is a straight line. Ray OX stands on it.
∴ b + c = 180º (Linear Pair)
54º + c = 180º
c = 180º − 54º = 126º
∴ c = 126º
In Figure, ∠ PQR = ∠ PRQ, then prove that∠ PQS = ∠ PRT.
In the given figure, ST is a straight line and ray QP stands on it.
∴ ∠PQS + ∠PQR = 180º (Linear Pair)
∠PQR = 180º − ∠PQS (1)
∠PRT + ∠PRQ = 180º (Linear Pair)
∠PRQ = 180º − ∠PRT (2)
It is given that ∠PQR = ∠PRQ.
Equating equations (1) and (2), we obtain
180º − ∠PQS = 180
− ∠PRT
∠PQS = ∠PRT
In Fig, if x + y = w + z, then prove that AOB is a line.
It can be observed that,
x + y + z + w = 360º (Complete angle)
It is given that,
x + y = z + w
∴ x + y + x + y = 360º
2(x + y) = 360º
x + y = 180º
Since x and y form a linear pair, AOB is a line.
In Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ ROS = (∠ QOS – ∠ POS).
It is given that OR ⊥ PQ
∴ ∠POR = 90º
⇒ ∠POS + ∠SOR = 90º
∠ROS = 90º − ∠POS … (1)
∠QOR = 90º (As OR ⊥ PQ)
∠QOS − ∠ROS = 90º
∠ROS = ∠QOS − 90º … (2)
On adding equations (1) and (2), we obtain
2 ∠ROS = ∠QOS − ∠POS
∠ROS = 
(∠QOS − ∠POS)
It is given that ∠ XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP, find ∠ XYQ and reflex ∠ QYP.
It is given that line YQ bisects ∠PYZ.
Hence, ∠QYP = ∠ZYQ
It can be observed that PX is a line. Rays YQ and YZ stand on it.
∴ ∠XYZ + ∠ZYQ + ∠QYP = 180º
⇒ 64º + 2∠QYP = 180º
⇒ 2∠QYP = 180º − 64º = 116º
⇒ ∠QYP = 58º
Also, ∠ZYQ = ∠QYP = 58º
Reflex ∠QYP = 360º − 58º = 302º
∠XYQ = ∠XYZ + ∠ZYQ
= 64º + 58º = 122º
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