Exercise: 6.1 (Page No: 96)
1. In Fig. 6.13, lines AB and CD intersect at O. If AOC +BOE = 70° and BOD = 40°, find BOE and reflex COE.
Solution:
From the diagram, we have
(AOC +BOE +COE) and (COE +BOD +BOE) forms a straight line.
So, AOC+BOE +COE = COE +BOD+BOE = 180°
Now, by putting the values of AOC+BOE = 70° and BOD = 40° we get
COE = 110° and
BOE = 30°
2. In Fig. 6.14, lines XY and MN intersect at O. If POY = 90° and a : b = 2 : 3, find c.
Solution:
We know that the sum of linear pair are always equal to 180°
So,
POY +a +b = 180°
Putting the value of POY = 90° (as given in the question) we get,
a+b = 90°
Now, it is given that a : b = 2 : 3 so,
Let a be 2x and b be 3x
∴ 2x+3x = 90°
Solving this we get
5x = 90°
So, x = 18°
∴ a = 2×18° = 36°
Similarly, b can be calculated and the value will be
b = 3×18° = 54°
From the diagram, b+c also forms a straight angle so,
b+c = 180°
c+54° = 180°
∴ c = 126°
3. In Fig. 6.15, PQR = PRQ, then prove that PQS = PRT.
Solution:
Since ST is a straight line so,
∠PQS+∠PQR = 180° (linear pair) and
∠PRT+∠PRQ = 180° (linear pair)
Now, ∠PQS + ∠PQR = ∠PRT+∠PRQ = 180°
Since ∠PQR =∠PRQ (as given in the question)∠PQS = ∠PRT. (Hence proved)
4. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.
Solution:
For proving AOB is a straight line, we will have to prove x+y is a linear pair
i.e. x+y = 180°
We know that the angles around a point are 360° so,
x+y+w+z = 360°
In the question, it is given that,
x+y = w+z
So, (x+y)+(x+y) = 360°
2(x+y) = 360°
∴ (x+y) = 180° (Hence proved).
5. 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. Prove that ROS = ½ (QOS – POS).
Solution:
In the question, it is given that (OR ⊥ PQ) and POQ = 180°
So, POS+ROS+ROQ = 180°
Now, POS+ROS = 180°- 90° (Since POR = ROQ = 90°)
∴ POS + ROS = 90°
Now, QOS = ROQ+ROS
It is given that ROQ = 90°,
∴ QOS = 90° +ROS
Or, QOS – ROS = 90°
As POS + ROS = 90° and QOS – ROS = 90°, we get
POS + ROS = QOS – ROS
2 ROS + POS = QOS
Or, ROS = ½ (QOS – POS) (Hence proved).
6. 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.
Solution:
Here, XP is a straight line
So, XYZ +ZYP = 180°
Putting the value of XYZ = 64° we get,
64° +ZYP = 180°
∴ ZYP = 116°
From the diagram, we also know that ZYP = ZYQ + QYP
Now, as YQ bisects ZYP,
ZYQ = QYP
Or, ZYP = 2ZYQ
∴ ZYQ = QYP = 58°
Again, XYQ = XYZ + ZYQ
By putting the value of XYZ = 64° and ZYQ = 58° we get.
XYQ = 64°+58°
Or, XYQ = 122°
Now, reflex QYP = 180°+XYQ
We computed that the value of XYQ = 122°.
So,
QYP = 180°+122°
∴ QYP = 302°