3 Marks Question

Question 13 Marks

In the given figure, m and it are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.

Answer

Let the normal to mirrors m and n intersect at P.
Now, $\text{OB }\bot\text{ m }\bot\text{ OC},\ \bot\ \text{n}$ and $\text{m }\bot\text{ n}.$
$\Rightarrow\text{OB }\bot\text{ OC}$
$\Rightarrow\angle\text{APB}=90^\circ$
$\Rightarrow\angle2+\angle3=90^\circ$ (sum of acute angles of a right triangle is 90°)
By the laws of reflection, we have
$\angle1=\angle2$ and $\angle4=\angle3$ (angle of incidence = angle of reflection)
$\Rightarrow\angle1+\angle4=\angle2+\angle3=90^\circ$
$\Rightarrow\angle1+\angle2+\angle3+\angle4=180^\circ$
$\Rightarrow\angle\text{CAB}+\angle\text{ABD}=180^\circ$
But, $\angle\text{CAB}$ and $\angle\text{ABD}$ consecutive interior angles formed, when the transversal AB cuts CA and BD.
$\therefore$ CA || BD

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Question 23 Marks

Find the value of x for which the angles (2x - 5)º and (x - 10)º are the complementary angle.

Answer

two angles whose sum is 90º are called complementary angles.
It is given that the angles (2x - 5)º and (x - 10)º are the complementary angles.
$\therefore$ (2x - 5)º + (x - 10)º = 90º
⇒ 3xº - 15º = 90º
⇒ 3xº = 90º + 15º = 105º
$\Rightarrow\text{x}^\circ=\frac{105^\circ}{3}=35^\circ$
Thus, the value of x is 35.

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Question 33 Marks

If two straight lines intersect each other, then prove that the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

Answer

Let AB and CD be the two lines intersecting at a point O and let ray OE bisect $\angle\text{AOC}.$

Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let $\angle\text{COE}=1,\angle\text{AOE}=2,\angle\text{BOF}=3$ and $\angle\text{DOF}=4.$
We know that vertically-opposite angles are equal.
$\therefore\angle1=\angle4$ and $\angle2=\angle3$
But, $\angle1=\angle2$ $[$Since OE bisects $\angle\text{AOC}]$
$\therefore\angle4=\angle3$
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

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Question 43 Marks

In the adjoining figure, AOB a straight line. Find the value of x. Also find $\angle\text{AOC}$ and $\angle\text{BOD}.$

Answer

As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180º.
Therefore,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
⇒ (3x - 6)º + 55º + (x + 20)º = 180
⇒ 4x = 111º
⇒ x = 27.5º
Hence,
$\angle\text{AOC}=3\text{x}-6$
= 3 × 27.5 - 6
= 76.5º
and $\angle\text{BOD}-\text{x}+20$
= 27.5 + 20
= 47.5º

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Question 53 Marks

Find the angle whose supplement is four times its complement.

Answer

Let the measure of the required angle be xº.
Then, measure of its complement = (90 - x)º.
And, measure of its supplement = (180 - x)º.
(180 - x) = 4 (90 - x)
⇒ 180 - x = 360 - 4x
⇒ 3x = 180
⇒ x = 60
Hence, the measure of the required angle is 60º.

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Question 63 Marks

Find the measure of an angle which is 30º less than its supplement.

Answer

Let the measure of the angle be xº.
$\therefore$ supplement of = 180º - xº
It is given that,
(180º - xº) - xº = 30º
⇒ 180º - 2xº = 30º
⇒ 2xº = 180º - 30º
⇒ 2xº = 150º
⇒ xº = 75º
Thus, the measure of the angle is 75º.

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Question 73 Marks

In the adjoining figure, AOB is a straight line. Find the value of x. Also, find $\angle\text{AOC},\angle\text{COD}$ and $\angle\text{BOD}.$

Answer

AOB is a straight line.
Therefore,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
⇒ (3x + 7)º + (2x - 19)º + xº = 180º
⇒ 6x = 192º
⇒ x = 32º
Therefore,
$\angle\text{AOC}=3\times32^\circ+7=103^\circ$
$\angle\text{COD}=2\times32^\circ-19=45^\circ\text{ and}$
$\angle\text{BOD}=32^\circ$

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Question 83 Marks

In the given figure, AB || CD transversal t cuts them at E and F respectively. If EG and FG are the bisectors of $\angle\text{BEF}$ and $\angle\text{EFD}$ respectively, prove that $\angle\text{EGF}=90^\circ.$

Answer

AB || CD and a transversal t cuts them at E and F respectively.
$\Rightarrow\angle\text{BEF}+\angle\text{DFE}=180^\circ$ (interior angles)
$\Rightarrow\frac{1}{2}\angle\text{BEF}+\frac{1}{2}\angle\text{DFE}=90^\circ$
$\Rightarrow\angle\text{PGEF}+\angle\text{GFE}=90^\circ\ ....(\text{i)}$
Now, in $\triangle\text{GEF},$ by angle sum property
$\angle\text{GEF}+\angle\text{GFE}+\angle\text{EGF}=180^\circ$
$\Rightarrow90^\circ+\angle\text{EGF}=180^\circ$ ….[From (i)]
$\Rightarrow\angle\text{EGF}=90^\circ$

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Question 93 Marks

Find the angle whose complement is one-third of its supplement.

Answer

Let the measure of the required angle be xº.
Then, the measure of its complement = (90 - x)º.
And the measure of its supplement = (180 - x)º.
Therefore,
$(90 - \text{x}) = \frac{1}{3}(180-\text{x})$
⇒ 3(90 - x) = (180 - x)
⇒ 270 - 3x = 180 - x
⇒ 2x = 90
⇒ x = 45
Hence, the measure of the required angle is 45º.

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Question 103 Marks

In the adjoining figure, x. y. z = 5 : 4 : 6. If XOY is a straight line, find the values of x, y and z.

Answer

Let x = 5a, y = 4a and z = 6a
XOY is a straight line.
Therefore,
$\angle\text{XOP}+\angle\text{POQ}+\angle\text{YOQ}=180^\circ$
⇒ 5a + 4a + 6a = 180º
⇒ 5a = 180º
⇒ a = 12º
Therefore,
x ⇒ 5 × 12º = 60º
y ⇒ 4 × 12º = 48º
and z ⇒ 6 × 12º = 72º

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Question 113 Marks

In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of $\angle\text{AEF}$ and $\angle\text{EFD}$ respectively, prove that EP || FQ.

Answer

Since AB || CD and t is a transversal, we have
$\angle\text{AEF}=\angle\text{EFD}$ (alternate angles)
$\Rightarrow\frac{1}{2}\angle\text{AEF}=\frac{1}{2}\angle\text{EFD}$
$\Rightarrow\angle\text{PEF}=\angle\text{EFQ}$
But, these are alternate interior angles formed when the transversal EF cuts EP and FQ.
$\therefore$ EP || FQ

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Question 123 Marks

In the given figure, AB || CD. Find the value of x.

Answer

Since AB || CD and BC is a transversal.
So, $\angle\text{BCD}=\angle\text{ABC}=\text{x}^\circ$ [Alternate angles]
As BC || ED and CD is a transversal.
$\angle\text{BCD}+\angle\text{EDC}=180^\circ$
$\Rightarrow\angle\text{BCD}+75^\circ=180^\circ$
$\Rightarrow\angle\text{BCD}=180^\circ-75^\circ=105^\circ$
$\Rightarrow\angle\text{ABC}=105^\circ$ $[$ Since $\angle\text{BCD}=\angle\text{ABC}]$
$\therefore\text{x}^\circ=\angle\text{ABC}=105^\circ$
Hence, $\text{x}=105.$

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Maths 3 Marks Question

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