Exercise 6.1
- As per the figure below, lines, AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, you need to find what is ∠BOE and reflex ∠COE.
Solution:
By looking at the above figure, you know that-
(∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) will create a straight line.
Therefore, ∠AOC + ∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180°
Thus, now you need to put the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° you will derive
∠COE = 110° and ∠BOE = 30°
This, reflex ∠COE = 360° – 110° = 250°.
2. As per the figure below, you can see lines XY and MN intersect at O. If ∠POY = 90° and a: b = 2 : 3, what is c?
Solution:
You need to know that the sum of linear pair is always equal to 180°
Therefore,
∠POY + a +b = 180°
Now, put the value of ∠POY = 90° (provided in the question) you will get,
a+b = 90°
As per the question, you have the data that a : b = 2 : 3 hence,
Assume, a as 2x and b as 3x
\( \therefore \) 2x+3x = 90°
Solve this equation; you will get-
5x = 90°
Hence, x = 18°
\( \therefore \) a = 2\( \times \)18° = 36°
Likewise, you can calculate b, and its value will be
b = 3\( \times \)18° = 54°
As per the figure above, b+c also creates a straight angle, thus,
b+c = 180°
c+54° = 180°
Hence, c = 126°
3. As per the figure below, ∠PQR = ∠PRQ; you need to prove that ∠PQS = ∠PRT.
Solution:
As you can see that ST is a straight line, therefore,
∠PQS+∠PQR = 180° (linear pair) and
∠PRT + ∠PRQ = 180° (linear pair)
So, ∠PQS + ∠PQR = ∠PRT + ∠PRQ = 180°
As ∠PQR = ∠PRQ (as per the details in the question)
∠PQS = ∠PRT.
Hence proved.
4. As per the figure, if x+y = w+z, you need to prove that AOB is a line.
Solution:
If you want to prove that AOB is a straight line, you will first have to prove that x+y is a linear pair.
Which means x+y = 180°
You should know that the angles around a point are 360°, thus,
x+y+w+z = 360°
As per the question-
x+y = w+z
Therefore, (x+y)+(x+y) = 360°
2(x+y) = 360°
\( \therefore \) (x+y) = 180°.
5. As per the diagram below, POQ is a line. The Ray OR you see is perpendicular to the line PQ. There is another ray, OS lying between rays OP and OR. You need to prove that ∠ROS = ½ (∠QOS – ∠POS).
Solution:
As per the question, the information states that (OR ⊥ PQ) and ∠POQ = 180°
Therefore, ∠POS + ∠ROS + ∠ROQ = 180°
This means, ∠POS+∠ROS = 180°- 90° (As, ∠POR = ∠ROQ = 90°)
Therefore, ∠POS + ∠ROS = 90°
It makes, ∠QOS = ∠ROQ+∠ROS
As per the question, ∠ROQ = 90°,
\( \therefore \) ∠QOS = 90° + ∠ROS
Or you can also say that, ∠QOS – ∠ROS = 90°
Now, ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, this will you will get
∠POS + ∠ROS = ∠QOS – ∠ROS
2 ∠ROS + ∠POS = ∠QOS
Or, ∠ROS = ½ (∠QOS – ∠POS).
6. If ∠XYZ = 64° and XY is created to point P. You need to create a figure from this information. You need to assume that ray YQ bisects ∠ZYP; as per this scenario, you need to find ∠XYQ and reflex ∠QYP.
Solution:
As per the above diagram, you can see that XP is a straight line
Thus, ∠XYZ + ∠ZYP = 180°
Put the value of ∠XYZ = 64° you will get,
64° +∠ZYP = 180°
Therefore, ∠ZYP = 116°
As per the figure, you also get the information that ∠ZYP = ∠ZYQ + ∠QYP
So, as YQ bisects ∠ZYP,
∠ZYQ = ∠QYP
Or, ∠ZYP = 2∠ZYQ
Hence, ∠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
You computed earlier that the value of ∠XYQ = 122°.
Hence,
∠QYP = 180°+122°
\( \therefore \) ∠QYP = 302°
Exercise 6.2
- As per the diagram below, you need to find the values of x and y; after that, you are also required to show that AB II CD.
Solution:
After looking at the figure, you can make out that,
50°+ x= 180° (Linear Pair)
\( \Rightarrow \) x=130°………. (1)
Now, from the above diagram, it is pretty evident that y=130° (as vertically opposite angles are equivalent)
Also, you can see that x and y are alternate interior angles for lines AB and CD, so measures of these angles are equal.
Hence, line AB || CD.
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