3 Marks Question

Question 13 Marks

The exterior angles, obtained on producing the base of a triangle both ways are $104^\circ$ and $136^\circ$. Find all the angles of the triangle.

Answer

$\angle\text{ACD}=\angle\text{ABC}+\angle\text{BAC}$ [Exterior angle property]
Now $\angle\text{ABC}=180^\circ-136^\circ=44^\circ$ [Linera pair]
$\angle\text{ACB}=180^\circ-104^\circ=76^\circ$ [Linera pair]
Now, In $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ [Sum of all angles of a triangle]
$\Rightarrow\angle\text{A}+44^\circ+76^\circ=180^\circ$
$\Rightarrow\angle\text{A}+180^\circ-44^\circ-76^\circ$
$\Rightarrow\angle\text{A}=60^\circ$

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Question 23 Marks

In figure $AE$ bisects $\angle\text{CAD}$ and $\angle\text{B}=\angle\text{C}.$ Prove that $AE\ ||\ BC$.

Answer

Let $\angle\text{B}=\angle\text{C}=\text{x}$
Then, $\angle\text{CAD}=\angle\text{B}+\angle\text{C}=2\text{x}$ (exterior angle)
$\Rightarrow\frac{1}{2}\angle\text{CAD}=\text{x}$
$\Rightarrow\angle\text{EAC}=\text{x}$
$\Rightarrow\angle\text{EAC}=\angle\text{C}$
These are alternate interior angles for the lines $AE$ and $BC$
$\therefore\text{AE }||\text{ BC}$

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Question 33 Marks

In a $\triangle\text{ ABC},\text{ AD}$ bisects $\angle\text{A}$ and $\angle\text{C} > \angle\text{B}.$. Prove that $\angle\text{ADB} > \angle\text{ADC}.$

Answer

$\therefore\angle\text{C}>\angle\text{B}$ [Given]
$\Rightarrow\angle\text{C}+\text{x} > \angle\text{B}+\text{x}$ [Adding $x$ on both sides]
$\Rightarrow180^\circ-\angle\text{ADC}>180^\circ-\angle\text{ADB}$
$\Rightarrow-\angle\text{ADC}>-\angle\text{ADB}$
$\Rightarrow\angle\text{ADB}>\angle\text{ADC}$ hence proved.

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Question 43 Marks

In Fig. $\text{AC}\perp\text{CE}$ and $\angle\text{A}:\angle\text{B}:\angle\text{C}=3:2:1,$ find the value of $\angle\text{ECD}.$

Answer

$\angle\text{A}:\angle\text{B}:\angle\text{C}=3:2:1$
Let the angles be $3x, 2x$ and $x$
$\Rightarrow 3x + 2x + x = 180^\circ $ [Angle sum property]
$\Rightarrow 6x = 180^\circ $
$\Rightarrow\text{x}=30=\angle\text{ACB}$
$\therefore\angle\text{ECD}=180^\circ-\angle\text{ACB}-90^\circ$[Linear pair]
$=180^\circ-30^\circ-90^\circ$
$=60^\circ$
$\therefore\angle\text{ECD}=60^\circ$

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Question 53 Marks

Two angles of a triangle are equal and the third angle is greater than each of those angles by $30^\circ$. Determine all the angles of the triangle.

Answer

Given that, Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°.
Let $x, x, x + 30^\circ $ be the angles of a triangle We know that, Sum of all angles in a triangle is $180^\circ x + x + x + 30^\circ = 180^\circ\ 3x + 30^\circ = 180^\circ\ 3x = 180^\circ - 30^\circ\ 3x = 150^\circ\ x = 50^\circ $.
Therefore, the three angles are $50^\circ , 50^\circ , 80^\circ $.

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Question 63 Marks

In Fig. $AB\ ||\ DE$ Find $\angle\text{ACD}.$

Answer

Since $AB\ ||\ DE$ $\therefore\angle\text{ABC}=\angle\text{CDE}=40^\circ$ [Alternate angles]
$\therefore\angle\text{ACB}=180^\circ-\angle\text{ABC}-\angle\text{BAC}$
$=180^\circ-40^\circ-30^\circ$ $=110^\circ$
$\therefore\angle\text{ACD}=180^\circ-110^\circ$ [Linear pair] $=70^\circ$

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Question 73 Marks

If the angles of a triangle are in the ratio $1 : 2 : 3$, determine three angles.

Answer

Given Data: Angles of a triangle are in the ratio $1: 2: 3$ Let the angles be $x, 2x, 3x$.
$\therefore$ We know that, Sum of all angles of triangles is $180^\circ $ $\text{x}+2\text{x}+3\text{x}=180^\circ$
$\Rightarrow6\text{x}=180^\circ$
$\Rightarrow\text{x}=\frac{180^\circ}{6}$
$\Rightarrow\text{x}=30^\circ$
Since $\text{x}=30^\circ$ $2x = 2(30)^\circ = 60^\circ\ 3x = 3(30)^\circ = 90^\circ $
Therefore, angles are $30^\circ , 60^\circ , 90^\circ $.

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Question 83 Marks

In a $\triangle\text{ABC},$ if $\angle\text{A}=55^\circ,\angle\text{B}=40^\circ,$ find $\angle\text{C}.$

Answer

Given Data:
$\angle\text{B}=55^\circ,\angle\text{B}=40^\circ,$ then $\angle\text{C}=?$
We know that
In a $\triangle\text{ABC}$ sum of all angles of a triangle is 180°
i.e., $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow55^\circ+40^\circ+\angle\text{C}=180^\circ$
$\Rightarrow95^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-95^\circ$
$\Rightarrow\angle\text{C}=85^\circ$

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Question 93 Marks

The angles of a triangle are (x - 40)º, (x - 20)º and $\Big(\frac{1}{2}\text{x}-10\Big)^\circ.$ Find the value of x.

Answer

Given that,
The angles of a triangle are
$(\text{x}-40^\circ),(\text{x}-20^\circ)$ and $\Big(\frac{1}{2}\text{x}-10^\circ\Big)$
We know that,
Sum of all angles of triangle is 180°
$\therefore(\text{x}-40^\circ)+(\text{x}-20^\circ)+\Big(\frac{1}{2}\text{x}-10^\circ\Big)=180^\circ$
$2\text{x}+\frac{1}{2}\text{x}-70^\circ=180^\circ$
$\frac{5}{2}\text{x}=180^\circ+70^\circ$
$5\text{x}=2(250)^\circ$
$\text{x}=\frac{500^\circ}{5}$
$\therefore\text{x}=100^\circ$

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Question 103 Marks

In a $\triangle\text{ ABC},\text{ BD}\perp\text{AC}$ and $\text{ CE}\perp\text{AB}.$ If $BD$ and $CE$ intersect $O$, prove that $\angle\text{BOC}=180^\circ-\text{A}.$

Answer

In quadrilateral $AEOD$ $\angle\text{A}+\angle\text{AEO}+\angle\text{EOD}+\angle\text{ADO}=360^\circ$
$\Rightarrow\angle\text{A}+90^\circ+90^\circ+\angle\text{EOD}=360^\circ$
$\Rightarrow\angle\text{A}+\angle\text{BOC}=180^\circ$
$[\therefore\angle\text{EOD}=\angle\text{BOC}$ vertically opposite angles$]$
$\Rightarrow\angle\text{BOC}=180^\circ-\angle\text{A}$

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