4 Marks Questions

Question 14 Marks

In $\triangle\text{PQR},$ if $\angle\text{P}-\angle\text{Q}=42^\circ$ and $\angle\text{Q}+\angle\text{R}=21^\circ,$ find $\angle\text{P},\angle\text{Q}$ and $\angle\text{R}.$

Answer

Since, $\angle\text{P},\angle\text{Q}$ and $\angle\text{R}$ are the angles of a triangle.
So, $\angle\text{P}+\angle\text{Q}+\angle\text{R}=180^\circ\ ...(\text{i)}$
Now, $\angle\text{P}-\angle\text{Q}=42^\circ$ [Given]
$\Rightarrow\angle\text{P}=42^\circ+\angle\text{Q}\ ....(\text{ii)}$ and $\angle\text{Q}-\angle\text{R}=21^\circ$ [Given]
$\angle\text{R}=\angle\text{Q}-21^\circ\ ....(\text{iii)}$
Substituting the value of $\angle\text{P}$ and $\angle\text{R}$ from $(ii)$ and $(iii)$ in $(i)$, we get,
$42^\circ+\angle\text{Q}+\angle\text{Q}+\angle\text{Q}-21^\circ=180^\circ$
$\Rightarrow3\angle\text{Q}+21^\circ=180^\circ$
$\Rightarrow3\angle\text{Q}=180^\circ-21^\circ=159^\circ$
$\Rightarrow\angle\text{Q}=\frac{159^\circ}{3}=53^\circ$
$\therefore\angle\text{P}=42^\circ+\angle\text{Q}$ $=42^\circ+53^\circ=95^\circ$
$\angle\text{R}=\angle\text{Q}-21^\circ$ $=53^\circ-21^\circ=32^\circ$
$\therefore\angle\text{P}=95^\circ,\angle\text{Q}=53^\circ$ and $\angle\text{R}=32^\circ.$

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Question 24 Marks

In a $\triangle\text{ABC},$ it is given that $\angle\text{A}:\angle\text{B}:\angle\text{C}=3:2:1$ and $\text{CD}\perp\text{AC}.$ Find $\angle\text{ECD}.$

Answer

In the given $\triangle\text{ABC},$
we have, $\angle\text{A}:\angle\text{B}:\angle\text{C}=3:2:1$
Let $\angle\text{A}=3\text{x},\angle\text{B}=2\text{x},\angle\text{C}=\text{x}.$
Then, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow3\text{x}+2\text{x}+\text{x}=180^\circ$
$\Rightarrow6\text{x}=180^\circ$
$\Rightarrow\text{x}=30^\circ$
$\therefore\angle\text{A}=3\text{x}=3\times30^\circ=90^\circ$
$\angle\text{B}=2\text{x}=2\times30^\circ=60^\circ$ and $\angle\text{C}=\text{x}=30^\circ$
Now, in ABC, we have, Ext. $\angle\text{ACE}=\angle\text{A}+\angle\text{B}=90^\circ+60^\circ=150^\circ$
$\therefore\angle\text{ACD}+\angle\text{ECD}=150^\circ$
$\Rightarrow\angle\text{ECD}=150^\circ-\angle\text{ACD}$
$\Rightarrow\angle\text{ECD}=150^\circ-90^\circ$
$\big[\text{since}\ \text{AD}\perp\text{CD},\ \angle\text{ACD}=90^\circ\big]$
$\Rightarrow\angle\text{ECD}=60^\circ$

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Question 34 Marks

Calculate the value of $x$ in the following figures.

Answer

In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow30^\circ+40^\circ+\angle\text{C}=180^\circ$
$\Rightarrow70^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-70^\circ=110^\circ$
Now $\angle\text{BCA}+\angle\text{ACD}=180^\circ$ [Linear pair]
$\Rightarrow110^\circ+\angle\text{ACD}=180^\circ$
$\Rightarrow\angle\text{ACD}=180^\circ-110^\circ=70^\circ$ In $\triangle\text{ECD},$
$\angle\text{ECD}+\angle\text{CDE}+\angle\text{CED}=180^\circ$
$\Rightarrow70^\circ+50^\circ+\angle\text{CED}=180^\circ$
$\Rightarrow120^\circ+\angle\text{CED}=180^\circ$
$\Rightarrow\angle\text{CED}=180^\circ-120^\circ=60^\circ$
Since $\angle\text{AED}$ and $\angle\text{CED}$ from a linear pair So,
$\angle\text{AED}+\angle\text{CED}=180^\circ$
$\Rightarrow\text{x}^\circ+60^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-60^\circ=120^\circ$
$\therefore\text{x}=120$

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Question 44 Marks

In the figure given alongside, $AB || CD, EF || BC$, $\angle\text{BAC}=60^\circ$ and $\angle\text{DHE}=50^\circ.$ Find $\angle\text{GCH}$ and $\angle\text{AGH}.$

Answer

$AB || CD$ and $AC$ is the transversal.
$\Rightarrow\angle\text{BAC}=\angle\text{ACD}=60^\circ$ (alternate angles) i. e.
$\angle\text{BAC}=\angle\text{GCH}=60^\circ$
Now, $\angle\text{DHF}=\angle\text{CHG}=50^\circ$ (vertically opposite angles) In $\triangle\text{GCH},$
by angle sum property, $\angle\text{GCH}+\angle\text{CHG}+\angle\text{CGH}=180^\circ$
$\Rightarrow60^\circ+50^\circ\angle\text{CGH}=180^\circ$
$\Rightarrow\angle\text{CGH}=70^\circ$
Now, $\angle\text{CGH}+\angle\text{AGH}=180^\circ$ (linear pair)
$\Rightarrow70^\circ+\angle\text{AGH}=180^\circ$

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Question 54 Marks

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.

Answer

Given: $A$ $\triangle\text{ABC}$ in which $BC, CA$ and $AB$ are produced to $D, E$ and $F$ respectively.
To prove: Exterior $\angle\text{DCA}=\angle\text{A}+\angle\text{B}\ .....(\text{i)}$
Exterior $\angle\text{FAE}=\angle\text{B}+\angle\text{C}\ .....(\text{ii)}$
Exterior $\angle\text{FBD}=\angle\text{A}+\angle\text{C}\ .....(\text{iii)}$
Adding $(i), (ii)$ and $(iii)$, we get,
$\text{Ext}.\angle\text{DCA}+\text{Ext}.\angle\text{FAE}+\text{Ext}.\angle\text{FBD}$ $=\angle\text{A}+\angle\text{B}+\angle\text{B}+\angle\text{C}+\angle\text{A}+\angle\text{C}$
$=2\angle\text{A}+2\angle\text{B}+2\angle\text{C}$
$=2(\angle\text{A}+\angle\text{B}+\angle\text{C})$
$=2\times180^\circ$ $[$Since, in triangle the sum of all three angle is $180^\circ ] = 360^\circ $
Hence, proved.

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Question 64 Marks

In the given figure, $\text{AM}\perp\text{BC}$ and $AN$ is the bisector of $\angle\text{A}.$ If $\angle\text{ABC}=70^\circ$ and $\angle\text{ACB}=20^\circ,$ find $\angle\text{MAN}.$

Answer

In $\triangle\text{ABC},$ by angle sum property,
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+70^\circ+20^\circ=180^\circ$
$\Rightarrow\angle\text{A}=90^\circ$
In $\triangle\text{ABM},$ by angle sum property,
$\angle\text{BAM}+\angle\text{ABM}+\angle\text{AMB}=180^\circ$
$\Rightarrow\angle\text{BAM}+70^\circ+90^\circ=180^\circ$
$\Rightarrow\angle\text{BAM}=20^\circ$
Since $AN$ is the bisector of $\angle\text{A},$
$\angle\text{BAN}=\frac{1}{2}\angle\text{A}$
$\Rightarrow\angle\text{BAN}=\frac{1}{2}\times90^\circ=45^\circ$
Now, $\angle\text{MAN}+\angle\text{BAM}=\angle\text{BAN}$
$\Rightarrow\angle\text{MAN}+20^\circ=45^\circ$
$\Rightarrow\angle\text{MAN}=25^\circ$

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Question 74 Marks

If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse-angled.

Answer

Let $ABC$ be a triangle and $\angle\text{B}>\angle\text{A}+\angle\text{C}$
Since, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+\angle\text{C}=180^\circ-\angle\text{B}$
Therefore, we get,
$\angle\text{B}>180^\circ-\angle\text{B}$
Adding $\angle\text{B}$ on both sides of the inequality, we get,
$\angle\text{B}+\angle\text{B}>180^\circ-\angle\text{B}+\angle\text{B}$
$\Rightarrow2\angle\text{B}>180^\circ$
$\Rightarrow\angle\text{B}>\frac{180^\circ}{2}=90^\circ$
i. e., $\angle\text{B}>90^\circ$ which means $\angle\text{B}$ is an obtuse angle.
$\therefore\triangle\text{ABC}$ is an obtuse angled triangle.

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Question 84 Marks

In the given figure, $AB || DE$ and $BD || FG$ such that $\angle\text{ABC}=50^\circ$ and $\angle\text{FGH}=120^\circ.$ Find the value of $x$ and $y$.

Answer

$\angle\text{FGH}+\angle\text{FGE}=180^\circ$ (linear pair)
$\Rightarrow120^\circ+\text{y}=180^\circ$
$\Rightarrow\text{y}=60^\circ$ $AB || DF$ and $BD$ is the transversal.
$\Rightarrow\angle\text{CDE}=50^\circ$ $BD || FG$ and $DF$ is the transversal.
$\Rightarrow\angle\text{EFG}=\angle\text{CDE}$ (alternate angles)
$\Rightarrow\angle\text{EFD}=50^\circ$ In $\triangle\text{EFG},$ by angle sum property,
$\angle\text{FEG}+\angle\text{FGE}+\angle\text{EFG}=180^\circ$
$\Rightarrow\text{x}+\text{y}+50^\circ=180^\circ$
$\Rightarrow\text{x}+60^\circ+50^\circ=180^\circ$
$\Rightarrow\text{x}=70^\circ$

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