3 Mark Question

Question 13 Marks

Two angle of a triangle are equal and the third angle is greater than each one of them by $18^{\circ}$. Find the angles.

Answer

Let the two equal angles, $\angle A$ and $\angle B$, of the triangle be $x ^{\circ}$ each. We know, $\angle A +\angle B +\angle C =180^{\circ}$
$\Rightarrow\text{x}^\circ+\text{x}^\circ+\angle\text{C}=180^\circ$
$\Rightarrow2\text{x}^\circ+\angle\text{C}=180^\circ\ ...(\text{i)}$
Also, it is given that,$\angle\text{C}=\text{x}^\circ+18^\circ\ ....(\text{ii)}$
Substituting $\angle\text{C}$ from (ii) in (i), we get,$2\text{x}^\circ+\text{x}^\circ+18^\circ=180^\circ$
$\Rightarrow3\text{x}^\circ=180^\circ-18^\circ=162^\circ$
$\Rightarrow\text{x}=\frac{162}{3}=54$
Thus, the required angles of the triangle are $54^{\circ}, 54^{\circ}$ and $x^{\circ}+18^{\circ}=54^{\circ}+18^{\circ}=72^{\circ}$.

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Question 23 Marks

In a right-angled triangle, one of the acute angles measures $53^{\circ}$. Find the measure of each angle of the triangle.

Answer

Let ABC be a right angled triangle and $\angle\text{C}=90^\circ$ Since,
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+\angle\text{B}=180^\circ-\angle\text{C}=180^\circ-90^\circ=90^\circ$
Suppose $\angle\text{A}=53^\circ$ Then, $53^\circ+\angle\text{B}=90^\circ$
$\Rightarrow\angle\text{B}=90^\circ-53^\circ=37^\circ$
$\therefore$ The required angles are $53^{\circ}, 37^{\circ}$ and $90^{\circ}$.

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Question 33 Marks

In the given figure, BAD || EF, $\angle\text{AEF}=55^\circ$ and $\angle\text{ACB}=25^\circ,$ find $\angle\text{ABC}.$

Answer

BAD || EF and EC is the transversal.$\Rightarrow\angle\text{AEF}=\angle\text{CAD}$ (corresponding angles)
$\Rightarrow\angle\text{CAD}=55^\circ$
Now, $\angle\text{CAD}+\angle\text{CAB}=180^\circ$ (linear pair)$\Rightarrow55^\circ+\angle\text{CAB}=180^\circ$
$\Rightarrow\angle\text{CAB}=125^\circ$
In $\triangle\text{ABC},$ by angle sum property,$\angle\text{ABC}+\angle\text{CAB}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ABC}+125^\circ+25^\circ=180^\circ$
$ \Rightarrow\angle\text{ABC}=30^\circ$

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Question 43 Marks

In the given figure, AB || CD and EF is a transversal. If $\angle\text{AEF}=65^\circ,\angle\text{DFG}=30^\circ,\angle\text{EFG}=90^\circ$
and $\angle\text{GEF}=\text{x}^\circ,$ find the value of x.

Answer

AB || CD and EF is the transversal.$\Rightarrow\angle\text{AEF}=\angle\text{EFD}$ (alternate angles)
$\Rightarrow\angle\text{AEF}=\angle\text{EFG}+\angle\text{DFG}$
$\Rightarrow65^\circ=\angle\text{EFG}+30^\circ$
$\Rightarrow\angle\text{EFG}=35^\circ$
In $\triangle\text{GEF},$ by angle sum property,$\angle\text{GEF}+\angle\text{EGF}+\angle\text{EFG}=180^\circ$
$\Rightarrow\text{x}+90^\circ+35^\circ=180^\circ$
$\Rightarrow\text{x}=55^\circ$

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Question 53 Marks

Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.

Answer

Let $\angle\text{C}$ be the smallest angle of $\angle\text{ABC}.$ Then, $\angle\text{A}=2\angle\text{C}$ and $\angle\text{B}=3\angle\text{C}$ Also, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$$\Rightarrow2\angle\text{C}+3\angle\text{C}+\angle\text{C}=180^\circ$
$\Rightarrow6\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=30^\circ$
So, $\angle\text{A}=2\angle\text{C}=2\times30^\circ=60^\circ$$\angle\text{B}=3\angle\text{C}=3\times30^\circ=90^\circ$
$\therefore$ The required angles of the triangle are 60º, 90º, 30º.

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Question 63 Marks

In $\triangle\text{ABC},$ if $\angle\text{A}+\angle\text{B}=108^\circ$ and $\angle\text{B}+\angle\text{C}=130^\circ,$ find $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$

Answer

$\angle\text{A}+\angle\text{B}=108^\circ$ [Given]But as $\angle\text{A},\angle\text{B}$ and $\angle\text{C}$ are the angles of a triangle,
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow108^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-108^\circ=72^\circ$
Also, $\angle\text{B}+\angle\text{C}=130^\circ$ [Given]
$\Rightarrow\angle\text{B}+72^\circ=130^\circ$
$\Rightarrow\angle\text{B}+72^\circ=130^\circ$
$\Rightarrow\angle\text{B}=130^\circ-72^\circ=58^\circ$
Now as, $\angle\text{A}+58^\circ=108^\circ$
$\Rightarrow\angle\text{A}+58^\circ=108^\circ$
$\Rightarrow\angle\text{A}=108^\circ-58^\circ=50^\circ$
$\therefore\angle\text{A}=50^\circ,\angle\text{B}=58^\circ$ and $\angle\text{C}=72^\circ$

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Question 73 Marks

The sum of two angles of a triangle is 116º and their differance is 24º. Find the measure of each angle of the triangle.

Answer

Given that the sum of the angles A and B of a $\triangle\text{ABC}$ is 116º, i. e., $\angle\text{A}+\angle\text{B}=116^\circ.$ Since, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ So, $116^\circ+\angle\text{C}=180^\circ$$\Rightarrow\angle\text{C}=180^\circ-116^\circ=64^\circ$
Also, it is given that:$\angle\text{A}-\angle\text{B}=24^\circ$
$\Rightarrow\angle\text{A}=24^\circ+\angle\text{B}$
Putting, $\angle\text{A}=24^\circ+\angle\text{B}$ in $\angle\text{A}+\angle\text{B}=116^\circ,$ we get,$24^\circ+\angle\text{B}+\angle\text{B}=116^\circ$
$\Rightarrow2\angle\text{B}+24^\circ=116^\circ$
$\Rightarrow2\angle\text{B}=116^\circ-24^\circ=92^\circ$
$\Rightarrow\angle\text{B}=\frac{92^\circ}{2}=46^\circ$
Therefore, $\angle\text{A}=24^\circ+46^\circ=70^\circ$$\therefore\angle\text{A}=70^\circ,\angle\text{B}=46^\circ$ and $\angle\text{C}=64^\circ.$

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Question 83 Marks

In the given figure, AB || CD, $\angle\text{BAE}=65^\circ$ and $\angle\text{OEC}=20^\circ.$ Find $\angle\text{ECO}.$

Answer

AB || CD and AE is the transversal.$\Rightarrow\angle\text{BAE}=\angle\text{DOE}$ (corresponding angles)
$\angle\text{DOE}=65^\circ$
Now, $\angle\text{DOE}+\angle\text{COE}=180^\circ$ (linear pair)$\Rightarrow65^\circ+\angle\text{COE}=180^\circ$
$\Rightarrow\angle\text{COE}=115^\circ$
In $\triangle\text{OCE},$ by angle sum property,$\angle\text{OEC}+\angle\text{ECO}+\angle\text{COE}=180^\circ$
$\Rightarrow20^\circ+\angle\text{ECO}+115^\circ=180^\circ$
$\Rightarrow\angle\text{ECO}=45^\circ$

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Question 93 Marks

The angles of a triangles are in the ratio $2: 3: 4$ Find the angles.

Answer

Let the measures of the angles of a triangle are $(2 x)^{\circ},(3 x)^{\circ}$ and $(4 x)^{\circ}$.
Then, $2 x+3 x+4 x=180$ [sum of the angles of a triangle is $180^{\circ}$ ]
$\Rightarrow 9 x=180$
$\Rightarrow x=\frac{180}{9}=20$
$\therefore$ The measures of the required angles are:
$2 x=(2 \times 20)^{\circ}$
$=40^{\circ} 3 x=(3 \times 20)^{\circ}$
$=60^{\circ} 4 x=(4 \times 20)^{\circ}=80^{\circ}$

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Question 103 Marks

If one angle of a triangle is equal to the sum of the other two, show that the triangle is right angled.

Answer

Let ABC be a triangle. Then,$\angle\text{A}=\angle\text{B}+\angle\text{C}$
$\therefore\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ [Sum of the angles of a triangle]
$\Rightarrow\angle\text {B}+\angle\text{C}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow2\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}=90^\circ$
$\Rightarrow\angle\text{A}=90^\circ$ $[\because\angle\text{A}=\angle\text{B}+\angle\text{C}]$
This implies that the triangle is right-angled at A.

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Question 113 Marks

In the given figure, AB || CD and EF is a transversal, cutting them at G and H respectively. If $\angle\text{EGB}=35^\circ$ and $\text{QP}\perp\text{EF},$ find the measure of $\angle\text{PQH}.$

Answer

AB || CD and EF is the transversal.$\Rightarrow\angle\text{EGB}=\angle\text{GHD}$ (corresponding angles)
$\angle\text{GHD}=35^\circ$
Now, $\angle\text{GHD}=\angle\text{QHP}$ (vertically opposite angles)$\Rightarrow\angle\text{QHP}=35^\circ$
In DQHP, by angle sum property,$\angle\text{PQH}+\angle\text{QHP}+\angle\text{QPH}=180^\circ$
$\Rightarrow\angle\text{PQH}+35^\circ+90^\circ=180^\circ$
$\Rightarrow\angle\text{PQH}=55^\circ$

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Question 123 Marks

In $\triangle\text{ABC},$ if $\angle\text{A}+\angle\text{B}=125^\circ$ and $\angle\text{A}+\angle\text{C}=113^\circ,$ find $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$

Answer

Since. $\angle\text{A},\angle\text{B}$ and $\angle\text{C}$ are the angles of a triangle . So, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ Now, $\angle\text{A}+\angle\text{B}=125^\circ$ [Given]$\therefore125^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ=125^\circ=55^\circ$
Also, $\angle\text{A}+\angle\text{C}=113^\circ$ [Given]$\Rightarrow\angle\text{A}+55^\circ=113^\circ$
$\Rightarrow\angle\text{A}=113^\circ-55^\circ=58 ^\circ$
Now as $\angle\text{A}+\angle\text{B}=125^\circ$$\Rightarrow58^\circ+\angle\text{B}=125^\circ$
$\Rightarrow\angle\text{B}=125^\circ-58^\circ=67^\circ$
$\therefore\angle\text{A}=58^\circ,\angle\text{B}=67^\circ$ and $\angle\text{C}=55^\circ.$

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Question 133 Marks

Calculate the value of x in the following figures.

Answer

Since AB || CD and AD is a transversal. So, $\angle\text{BAD}=\angle\text{ADC}$$\Rightarrow\angle\text{ADC}=60^\circ$
In $\triangle\text{ECD},$ we have,$\angle\text{E}+\angle\text{C}+\angle\text{D}=180^\circ$
$\Rightarrow\text{x}^\circ+45^\circ+60^\circ=180^\circ$
$\Rightarrow\text{x}^\circ+105^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-105^\circ=75^\circ$
$\therefore\text{x}=75$

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Question 143 Marks

Calculate the value of x in the following figures.

Answer

$\angle\text{EAF}=\angle\text{BAC}$ [Vertically opposite angles]
$\Rightarrow\angle\text{BAC}=60^\circ$
In $\triangle\text{ABC},$ exterior $\angle\text{ACD}$ is equal to the sum of two opposite interior angles. So, $\angle\text{ACD}=\angle\text{BAC}+\angle\text{ABC}$$\Rightarrow115^\circ=60^\circ+\text{x}^\circ$
$\Rightarrow\text{x}^\circ=115^\circ-60^\circ=55^\circ$
$\therefore\text{x}=55$

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