We know that the sum of angles around a point is 360°. Therefore x + y + w + z = 360° ⇒ (x + y) + (w + z) = 360° ⇒ (x + y) + (w + z) =360° ⇒ 2(x + y) = 360° ⇒ (x + y) = 360°/2 ⇒ x + y = 180° ∠AOC and ∠COB are forming linear pair. Hence, AOB is a straight line.
We know that the sum of angles around a point is 360°. Therefore x + y + w + z = 360°
⇒ (x + y) + (w + z) = 360°
⇒ (x + y) + (w + z) =360°
⇒ 2(x + y) = 360°
⇒ (x + y) = 360°/2
⇒ x + y = 180°
∠AOC and ∠COB are forming linear pair. Hence, AOB is a straight line.
It is given that ∠ XYZ = 64° and XY is produced to point P.
∠PYZ + ∠XYZ = 180° [∵ Linear Pair] ⇒ ∠PYZ + 64° = 180° [∵ ∠XYZ = 64°] ⇒ ∠PYZ = 180° - 64° = 116° But, ∠PYQ = ∠ZYQ = 1/2 ∠PYZ [∵ ∠ZYP is bisected by ray YQ] ∴ ∠PYQ = ∠ZYQ = 1/2 × 116° = 58° ∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122 and Reflex ∠QYP = 360° - ∠QYP = 360° - 58° = 302°
∠PYZ + ∠XYZ = 180° [∵ Linear Pair]
See less⇒ ∠PYZ + 64° = 180° [∵ ∠XYZ = 64°]
⇒ ∠PYZ = 180° – 64° = 116°
But,
∠PYQ = ∠ZYQ = 1/2 ∠PYZ [∵ ∠ZYP is bisected by ray YQ]
∴ ∠PYQ = ∠ZYQ = 1/2 × 116° = 58°
∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122 and
Reflex ∠QYP = 360° – ∠QYP = 360° – 58° = 302°
In Fig. 6.15, ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.
∠PQS + ∠PQR = 180° [∵ Linear Pair] ∠PRQ + ∠PRT = 180° [∵ Linear Pair] But, ∠PQR = ∠PRQ ∴ From the equations (1) and (2), we have ∠PQS = ∠PRT
∠PQS + ∠PQR = 180° [∵ Linear Pair]
See less∠PRQ + ∠PRT = 180° [∵ Linear Pair]
But, ∠PQR = ∠PRQ
∴ From the equations (1) and (2), we have
∠PQS = ∠PRT
In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
We know that the sum of angles around a point is 360°. Therefore x + y + w + z = 360° ⇒ (x + y) + (w + z) = 360° ⇒ (x + y) + (w + z) =360° ⇒ 2(x + y) = 360° ⇒ (x + y) = 360°/2 ⇒ x + y = 180° ∠AOC and ∠COB are forming linear pair. Hence, AOB is a straight line.
We know that the sum of angles around a point is 360°. Therefore x + y + w + z = 360°
See less⇒ (x + y) + (w + z) = 360°
⇒ (x + y) + (w + z) =360°
⇒ 2(x + y) = 360°
⇒ (x + y) = 360°/2
⇒ x + y = 180°
∠AOC and ∠COB are forming linear pair. Hence, AOB is a straight line.
In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR.
RHS = 1/2 (∠QOS - ∠POS) = 1/2 [(∠QOP + ∠ROS) - (∠POR - ∠ROS)] [∵ ∠QOS = ∠QOR + ∠ROS and ∠POS = ∠POR - ∠ROS] = 1/2 [∠QOR + ∠ROS - ∠POR + ∠ROS] = 1/2 [90° + ∠ROS - 90° + ∠ROS] [∵ ∠QOR = 90° and ∠POR = 90°] = 1/2 [2∠ROS] = ∠ROS = LHS
RHS = 1/2 (∠QOS – ∠POS)
See less= 1/2 [(∠QOP + ∠ROS) – (∠POR – ∠ROS)]
[∵ ∠QOS = ∠QOR + ∠ROS and ∠POS = ∠POR – ∠ROS]
= 1/2 [∠QOR + ∠ROS – ∠POR + ∠ROS]
= 1/2 [90° + ∠ROS – 90° + ∠ROS] [∵ ∠QOR = 90° and ∠POR = 90°]
= 1/2 [2∠ROS]
= ∠ROS = LHS