3 Mark Question

Question 13 Marks

In Figure, AB || CD and AC || BD. Find the values of x, y, z.

Answer

Since, AC || BD and CD || AB, ABCD is a parallelogram
Adjacent angles of parallelogram,
$\angle \text{CAD}+\angle \text{ACD}=180^\circ$
$=\angle \text{ACD}=180^\circ-65^\circ$
$=115^\circ$
Opposite angles of parallelogram,
$=\angle \text{CAD}=\angle \text{CDB}=65^\circ$
$=\angle \text{ACD}=\angle \text{DBA}=115^\circ$

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Question 23 Marks

In Fig., it being given that $\angle1 = 65^\circ,$ find all the other angles.

Answer

Given,
$\angle1=\angle3 $ are the vertically opposite angles
Therefore, $\angle3 = 65^\circ$
Here, $\angle1 + \angle2 = 180^\circ$ are the linear pair
Therefore, $\angle2 = 180^\circ-65^\circ$
$= 115^\circ$
$\angle2 = \angle4$ are the vertically opposite angles
Therefore, $\angle2 = \angle4 = 115^\circ$
And $\angle3 = 65^\circ$

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Question 33 Marks

In Figure, find $\angle \text{x}.$ Further find $\angle \text{BOC}, \angle \text{COD}, \text{ and }\angle \text{AOD}$

Answer

$\angle \text{AOD}+\angle \text{DOC}+\angle \text{COB}=180^\circ$ (Linear pair)
$(\text{x}+10)^\circ+\text{x}^\circ+(\text{x}+20)^\circ=180^\circ$
$3\text{x}+30^\circ=180^\circ$
$3\text{x}=180^\circ-30^\circ$
$3\text{x}=150^\circ$
$\text{x}=\frac{150^\circ}{3}=50^\circ$
$\angle \text{BOC}=\text{x}+20^\circ$
$=50^\circ+20^\circ=70^\circ$
$\angle \text{COD}=\text{x}=50^\circ$
$\angle \text{AOD}=\text{x}+10^\circ$
$=50^\circ+10^\circ=60^\circ$

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Question 43 Marks

In Figure, OE is the bisector of $\angle \text{BOD}.$ If $\angle1 =70^\circ,$ Find the magnitude of $\angle2, \angle3, \angle4$

Answer

Since OE is the bisector of $\angle\text{BOD},$
$\therefore\angle\text{DOE}=\angle\text{EOB}$
$\angle2+\angle1+\angle\text{EOB}=180^\circ$ (Linear pair)
$\angle2+\angle1=180^\circ$ $(\angle1=\angle\text{EOB})$
$\Rightarrow\angle2=180^\circ-2\angle1=180^\circ-2\times 70^\circ$
$\Rightarrow 180^\circ-140^\circ=40^\circ$
$\angle4=\angle2=40^\circ$ (Vertically opposite angles)
$\angle3=\angle\text{DOB}=\angle1+\angle\text{EOB}$ $[\angle3=\angle\text{DOB}$ (Vertically opposite)$]$
$=70^\circ+70^\circ$
$=140^\circ$

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Question 53 Marks

If two supplementary angles have equal measure, what is the measure of each angle?

Answer

Let p and q be the two supplementary angles that are equal
$\angle \text{p}=\angle \text{q}$
So,
$\angle \text{p}+\angle \text{q}=180^\circ$
$\Rightarrow\angle \text{p}+\angle \text{p}=180^\circ$
$\Rightarrow2\angle \text{p}=180^\circ$
$\Rightarrow \angle \text{p}=\frac{180^\circ}{2}$
$\Rightarrow \angle \text{p}=90^\circ$
Therefore, $\angle \text{p}=\angle \text{q}=90^\circ$

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Question 63 Marks

How many pairs of adjacent angles, in all, can you name in Figure?

Answer

There are 10 adjacent pairs,
$\angle \text{EOD}\text{ and }\angle \text{DOC}$
$\angle \text{COD}\text{ and }\angle \text{BOC}$
$\angle \text{COB}\text{ and }\angle \text{BOA}$
$\angle \text{AOB}\text{ and }\angle \text{BOD}$
$\angle \text{BOC}\text{ and }\angle \text{COE}$
$\angle \text{COD}\text{ and }\angle \text{COA}$
$\angle \text{DOE}\text{ and }\angle \text{DOB}$
$\angle \text{EOD}\text{ and }\angle \text{DOA}$
$\angle \text{EOC}\text{ and }\angle \text{AOC}$
$\angle \text{AOB}\text{ and }\angle \text{BOE}$

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Question 73 Marks

In Figure, write all pairs of adjacent angles and all the linear pairs.

Answer

Pairs of adjacent angles are:
$\angle\text{DOA and }\angle\text{DOC}$
$\angle \text{BOC}\text{ and }\angle \text{COD}$
$\angle \text{AOD}\text{ and } \angle \text{BOD}$
$\angle \text{AOC}\text{ and }\angle \text{BOC}$
Linear pairs:
$\angle \text{AOD}\text{ and }\angle \text{BOD}$
$\angle \text{AOC}\text{ and }\angle \text{BOC}$

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Question 83 Marks

In Fig. OA and OB are the opposite rays:
If x = 25°, what is the value of y?

Answer

$\angle \text{AOC}+\angle \text{BOC}=180^\circ-\text{Linear pair}$
$\Rightarrow 2\text{y}+5+3\text{x}=180^\circ$
$\Rightarrow 3\text{x}+2\text{y}=175^\circ$
If $\text{x}=25^\circ,$ then
$\Rightarrow 3(25^\circ)+2\text{y}=175^\circ$
$\Rightarrow 75^\circ+2\text{y}=175^\circ$
$\Rightarrow 2\text{y}=175^\circ-75^\circ$
$\Rightarrow2\text{y}=100^\circ$
$\Rightarrow \text{y}=\frac{100^\circ}{2}$
$\Rightarrow \text{y}=50^\circ$

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Question 93 Marks

In Figure, name each linear pair and each pair of vertically opposite angles.

Answer

Two adjacent angles are said to be linear pair of angles, if their non-common arms are two opposite rays.
$\angle1 \text{ and } \angle2$
$\angle2 \text{ and } \angle3$
$\angle3 \text{ and } \angle4$
$\angle1 \text{ and } \angle4$
$\angle5 \text{ and } \angle6$
$\angle6 \text{ and } \angle7$
$\angle7 \text{ and } \angle8$
$\angle8 \text{ and } \angle5$
$\angle9 \text{ and } \angle10$
$\angle10 \text{ and } \angle11$
$\angle11 \text{ and } \angle12$
$\angle12 \text{ and } \angle9$
The two angles are said to be vertically opposite angles if the two intersecting lines have no common arms.
$\angle1 \text{ and } \angle3$
$\angle4 \text{ and } \angle2$
$\angle5 \text{ and } \angle7$
$\angle6 \text{ and } \angle8$
$\angle9 \text{ and } \angle11$
$\angle10 \text{ and } \angle12$

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Question 103 Marks

In Fig. OA and OB are the opposite rays:
If y = 35°, what is the value of x?

Answer

$\angle \text{AOC}+\angle \text{BOC}=180^\circ-\text{Linear pair}$
$\Rightarrow 2\text{y}+5+3\text{x}=180^\circ$
$\Rightarrow 3\text{x}+2\text{y}=175^\circ$
If $\text{y}=35^\circ,$ then
$3\text{x}+2(35^\circ)=175^\circ$
$\Rightarrow 3\text{x}+70^\circ=175^\circ$
$\Rightarrow 3\text{x}=175^\circ-70^\circ$
$\Rightarrow 3\text{x}=105^\circ$
$\Rightarrow \text{x}=\frac{105^\circ}{3}$
$\Rightarrow \text{x}=35^\circ$

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Question 113 Marks

If the complement of an angle is 28°, then find the supplement of the angle.

Answer

Here, let p be the complement of the given angle 28°
Therefore, $\angle \text{p}+28^\circ=90^\circ$
$\Rightarrow \angle \text{p}=90^\circ-28^\circ$
$=62^\circ$
So, the supplement of the angle = 180° - 62°
= 118°

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