5 Marks Questions

Question 15 Marks

In a $\triangle\text{ABC},$ the internal bisectors of $\angle\text{B}$ and $\angle\text{E}$ meet at P and the external bisectors of $\angle\text{B}$ and $\angle\text{C}$ meet at Q. Prove that $\angle\text{BPC}+\angle\text{BQC}=180^\circ.$

Answer

Let $\angle\text{ABC}=2\text{x}$ and $\angle\text{ACE}=2\text{y}$

$\angle\text{ABC}=180^\circ-2\text{x}$ [Linera pair]
$\angle\text{ACB}=180^\circ-2\text{y}$ [Linera pair]
$\angle\text{A}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ [Sum of all angles of a triangle]
$\Rightarrow\angle\text{A}+180^\circ-2\text{x}+180^\circ-2\text{y}=180^\circ$
$\Rightarrow-\angle\text{A}+2\text{x}+2\text{y}=180^\circ$
$\Rightarrow\text{x}+\text{y}=90^\circ+\frac{1}{2}\angle\text{A}$
Now in $\triangle\text{BQC}$
$\text{x}+\text{y}+\angle\text{BQC}=180^\circ$ [Sum of all angles of a triangle]
$\Rightarrow90^\circ+\frac{1}{2}\angle\text{A}+\angle\text{BQC}=180^\circ$
$\Rightarrow\angle\text{BQC}=90^\circ-\frac{1}{2}\angle\text{A}\dots(\text{i})$
and we know that $\Rightarrow\angle\text{BPC}=90^\circ+\frac{1}{2}\angle\text{A}\dots(\text{ii})$
Adding (i) and (ii) we get $\angle\text{BPC}+\angle\text{BQC}=180^\circ$
Hence proved.

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Question 25 Marks

In $\triangle\text{ABC},$ if bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ intersect at O at angle of 120°, then find the measure of $\angle\text{A}.$

Answer

In the given $\triangle\text{ABC},\angle\text{ABC}=\angle\text{ACB},$ the bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ meet at O and $\angle\text{BOC}=120^\circ$
We need to find the measure of $\angle\text{A}$

So here, using the corollary, "if the bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ of a $\triangle\text{ABC},$meet at a point O,
Then $\angle\text{BOC}=90^\circ+\frac{1}{2}\angle\text{A}"$
Thus, in $\triangle\text{ABC},$
$\angle\text{BOC}=90^\circ+\frac{1}{2}\angle\text{A}$
$120^\circ=90^\circ+\frac{1}{2}\angle\text{A}$
$120^\circ-90^\circ=\frac{1}{2}\angle\text{A}$
$\angle\text{A}=2(30^\circ)$
$\angle\text{A}=60^\circ$
Thus, $\angle\text{A}=60^\circ$

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Question 35 Marks

In Fig. $\text{AM}\perp\text{BC}$ and AN is the bisector of $\angle\text{A}.$ If $\angle\text{B}=65^\circ$ and $\angle\text{C}=33^\circ,$ find $\angle\text{MAN}.$

Answer

Let $\angle\text{BAN}=\angle\text{NAC}=\text{x} $ $[\therefore\text{AN}\text{ bisects }\angle\text{A}]$
$\therefore\angle\text{ANM}=\text{x}+33^\circ$ [Exterior angle property]
In $\triangle\text{AMB}$
$\angle\text{BAM}=90^\circ-65^\circ=25^\circ$ [Exterior angle property]
$\therefore\angle\text{MAN}=\angle\text{BAN}-\angle\text{BAM}=(\text{x}-25)^\circ$
Now in $\triangle\text{MAN},$
$(\text{x}-25)^\circ+(\text{x}+33)^\circ+90^\circ=180^\circ$ [Angle sum property]
$\Rightarrow2\text{x}+8^\circ=90^\circ$
$\Rightarrow2\text{x}=82^\circ$
$\Rightarrow\text{x}=41^\circ$
$\therefore\text{MAN}=\text{x}-25^\circ$
$=41^\circ-25^\circ$
$=16^\circ$

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Question 45 Marks

ABC is a triangle. The bisector of the exterior angle at B and the bisector of $\angle\text{C}$ intersect each other at D. Prove that $\angle\text{D}=\frac{1}{2}\angle\text{A}.$

Answer

Let $\angle\text{ABC}=2\text{x}$ and $\angle\text{ACB}=2\text{y}$

$\angle\text{ABC}=180^\circ-2\text{x}$ [Linear pair]
$\therefore\angle\text{A}=180^\circ-\angle\text{ABC}-\angle\text{ACB}$ [Angle sum property]
$=180^\circ-180^\circ+2\text{x}+2\text{y}$
$=2(\text{x}-\text{y})\dots(\text{i})$
Now, $\angle\text{D}=180^\circ-\angle\text{DBC}-\angle\text{DCB}$
$\Rightarrow\angle\text{D}=180^\circ-(\text{x}+180^\circ-2\text{x})-\text{y})$
$\Rightarrow\angle\text{D}=180^\circ-\text{x}-180^\circ+2\text{x}-\text{y})$
$=(\text{x}-\text{y})$
$=\frac{1}{2}\angle\text{A}\dots\text{from(i)}$
Hence, $\angle\text{D}=\frac{1}{2}\angle\text{A}.$

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Question 55 Marks

In the given figure, compute the value of x.

Answer

In the given figure, $\angle\text{DCB}=45^\circ,\angle\text{CBA}=35^\circ$ and $\angle\text{BAD}=35^\circ$
Here, we will produce AD to meet BC at E

Now, using angle sum property of the triangle
In $\triangle\text{AEB}$
$\angle\text{BAE}+\angle\text{AEB}+\angle\text{EBA}=180^\circ$
$\angle\text{AEB}+35^\circ+45^\circ=180^\circ$
$\angle\text{AEB}+180^\circ=180^\circ$
$\angle\text{AEB}=180^\circ-80^\circ$
$\angle\text{AEB}=100^\circ$
Eurther, BEC is a straight line. So, using the property, "the angles forming a linear pair are supplementary", we get
$\angle\text{AEB}+\angle\text{AEC}=180^\circ$
$100^\circ+\angle\text{AEC}=180^\circ$
$\angle\text{AEC}=180^\circ-100^\circ$
$\angle\text{AEC}=80^\circ$
Also, using the property, " an exterior angle of a triangle is equal to the sum of its two opposite interior angles"
In $\triangle\text{DEC},\text{x}$ is its exterior angle
Thus,
$\angle\text{x}=\angle\text{DCE}+\angle\text{DEC}$
$=50^\circ+80^\circ$
$=130^\circ$
Therefore, $\text{x}=130^\circ.$

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Question 65 Marks

In a $\triangle\text{ABC},\angle\text{ABC}=\angle\text{ACB}$ and the bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ intersect at O such that $\angle\text{BOC}=120^\circ.$ Show that $\angle\text{A}=\angle\text{B}=\angle\text{C}=60^\circ.$

Answer

Given,

In $\triangle\text{ABC},$
$\angle\text{ABC}=\angle\text{ACB}$
Dividing both sides by '2'
$\frac{1}{2}\angle\text{ABC}=\frac{1}{2}\angle\text{ACB}$
$\Rightarrow\angle\text{OBC}=\angle\text{OCB}$ $[\therefore\text{OB},\text{OC }\text{bisects }\angle\text{B}\text{ and }\angle\text{C}]$
Now,
$\angle\text{BOC}=90^\circ+\frac{1}{2}\angle\text{A}$
$\Rightarrow120^\circ-90^\circ=\frac{1}{2}\angle\text{A}$
$\Rightarrow30^\circ\times(2)=\angle\text{A}$
$\Rightarrow\angle\text{A}=60^\circ$
Now in $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ (Sum of all angle of a triangle)
$60^\circ+2\angle\text{ABC}=180^\circ$ $[\therefore\angle\text{ABC}=\angle\text{ACB}]$
$\Rightarrow2\angle\text{ABC}=180^\circ-60^\circ$
$\Rightarrow\angle\text{ABC}=\frac{120^\circ}{2}=60^\circ$
$\Rightarrow\angle\text{ABC}=\angle\text{ACB}$
$\therefore\angle\text{ACB}=60^\circ$
Hence Proved.

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Question 75 Marks

In Fig. the sides BC, CA and AB of a triangle ABC have been produced to D, E and F respectively. If $\angle\text{ACD}=105^\circ$ and $\angle\text{EAF}=45^\circ,$ find all the angles of the triangle ABC.

Answer

$\angle\text{BAC}=\angle\text{EAF}=45^\circ$ [Vertically opposite angles]
$\angle\text{ABC}=\angle\text{105}^\circ-45^\circ=60^\circ$ [Exterior angle property]
$\angle\text{ACD}=180^\circ-105^\circ=75^\circ$ [Linear pair]

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