1). In figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70⁰ and ∠BOD = 40⁰. Find ∠BOE and reflex ∠COE.
Given,
∠BOD = 40⁰ and ∠AOC + ∠BOE = 70⁰ ....(1)
Now, ∠BOD = ∠AOC = 40⁰ [ ∵ vertically opposite angle]
From (1),
< BOD + < BOE = 70⁰ [ put, < AOC = < BOD ]
Or, 40⁰ + < BOE = 70⁰ [ put, < BOD = 40⁰]
Or, < BOE = 70⁰ - 40⁰ = 30⁰
Or, < BOE = 30⁰
Now,
∠AOC + ∠COE + ∠BOE = 180⁰
or, ∠AOC + ∠BOE + ∠COE = 180
or, 70⁰ + ∠COE = 180⁰
or, ∠COE = 180 - 70 = 110⁰
Then, reflex ∠COE = 360⁰ - ∠COE
or, reflex ∠COE = 360 - 110 = 250⁰
Hence, ∠BOE = 30⁰ and reflex ∠COE = 250⁰
2). In figure, lines XY and MN intersect at O. If ∠POY = 90⁰ and a : b = 2 : 3, find c
Given,
∠POY = 90⁰
a : b = 2 : 3, then a = 2k and b = 3k
Now,
Ray OP stands on line XY
∴ ∠XOP + ∠POY = 180⁰ [[∵ ∠XOP = XOM + ∠POM = a + b]
or, a + b + 90⁰ = 180⁰
or, a + b = 90⁰ ………(1)
Substitute, a = 2k, b = 3k in (1)
∴ 2k + 3k = 90
or, 5k = 90⁰
or, k = 90/5 = 18⁰
∴ a = 2k = 2 × 18 = 36⁰
b = 3k = 3 × 18 = 54⁰
Also, ray OX stands on line MN
∴ ∠MOX + ∠XON = 180⁰
or, b + c = 180 [ ∵ XON = c]
or, 54 + c = 180 [ ∵ b = 54]
or, c = 180 - 54 = 126⁰
Hence, the value of c is 126⁰
3). In figure below, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Solution: Given, ∠PQR = ∠PRQ
Now, ray QP stands on a line ST, then
∠SQP + ∠PQR = 180⁰ [ ∵ Linear pair]….(1)
Also, ray RP stands on a line ST, then
∠QRP + ∠PRT = 180⁰ ……..(2) [ ∵ Linear pair]
Equating (1) and (2)
∠SQP + ∠PQR = ∠QRP + ∠PRT
or, ∠PQS + ∠PQR = ∠PRQ + ∠PRT
or, ∠PQS + ∠PQR = ∠PQR + ∠PRT
[ ∵ ∠PRQ = ∠PQR ]
or, ∠PQS = PRT proved.
4). In figure, if x + y = w + z, then prove that AOB is a line.
Solution : We know, sum of all the angles round a point is equal to 360⁰.
or, x + y + z + w = 360⁰
or, ( x + y) + ( z + w) = 360⁰ ……,(1)
But, x + y = z + w [ given]
Substitute, x + y = z + w in (1)
∴ ( x + y) + ( x + y) = 360
or, 2x + 2y = 360
or, 2( x + y) = 360
or, x + y = 360/2 = 180
or, x + y = 180⁰
or, AOB is a straight line.
5). 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 = 1/2( ∠QOS - ∠POS)
Solution: Given,
OR ⏊ PQ
∴ ∠POR = ∠QOR = 90⁰
or, ∠POR = ∠POS + ∠ROS = 90⁰ ……(1)
Ray OS stands on a line PQ
∴ ∠POS + ∠QOS = 180⁰ [ linear pair]
or, ∠POS + ∠QOS = 2.90⁰………(2)
Substitute, 90⁰ = ∠POS +∠ROS in (2)
∠POS + ∠QOS = 2 [ ∠POS + ∠ROS]
or, ∠POS + ∠QOS = 2∠POS + 2∠ROS
or, 2∠ROS = ∠QOS - ∠POS
or, ∠ROS = 1/2 [ ∠QOS - ∠POS] proved
6). It is given that ∠XYZ = 64⁰ and XY is produced to a point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Solution: Given, YQ bisects ∠ZYP.
Let ∠PYQ = ∠QYZ = x⁰
∴ ∠ PYZ = ∠PYQ + ∠QYZ = x + x = 2x
Now, ray YZ stands on a line PX, then
∴ ∠PYZ + ∠ZYX = 180⁰ [ Linear pair]
or,( x + x) + 64 = 180⁰
or, 2x + 64 = 180⁰
or, 2x = 180⁰ - 64⁰ = 116⁰
or, x = 116⁰/2 = 58⁰
∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64⁰ + 58⁰
or, ∠XYQ = 64 + 58 = 122⁰
and reflex∠QYP = 360⁰ - ∠QYP = 360-x
or, reflex ∠QYP = 360 - 58 = 302⁰
Hence, the figure is,
.and ∠XYQ = 122⁰ , reflex ∠QYP = 302⁰
