[5 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip

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Question 15 Marks

$\text{ABC}$ and $\text{DBC}$ are two isosceles triangles on the same side of $BC$. Prove that:$(i) DA$ or $(AD)$ produced bisects $BC$ at right angle.$(ii)\text{BDA} =\text{CDA}.$

Answer

$DA$ is produced to meet $BC$ in $L$.
In $\triangle A B C$
$ A B=A C \ldots[$ Given $]$
$\therefore \angle A C B=\angle A B C . \ldots \ldots .. ( i )\ldots[$ angles opposite to equal sides are equal $]$
In $\triangle DBC$
DB $=$ DC $ \ldots[$ Given $]$
$\therefore \angle DCB =\angle DBC . \ldots .. ( ii ) \ldots [$ angles opposite to equal sides are equal $]$
Subtracting $(i)$ from $(ii)$
$ \angle DCB -\angle ACB =\angle DBC -\angle ABC$
$ \Rightarrow \angle DCA =\angle DBA \ldots \ldots .( iii )$
In $\triangle DBA$ and $\triangle DCA$,
$DB = DC \ldots[$ Given $]$
$ \angle DBA =\angle DCA \ldots[$ From $( iii )]$
$ AB = AC \ldots[$ Given $]$
$ \therefore \triangle DBA \cong \triangle DCA \ldots .[ \text{SAS }]$
$ \Rightarrow \angle BDA =\angle CDA\dots . . . . . . \text { (iv ) ...[ c. p. c .t ] }$
In $\triangle DBA$,
$\angle BAL =\angle DBA +\angle BDA \ldots . . . ( v ) ...[$ Ext. angle $=$sum opp. int. angles$]$
From $(vi)$ and $(vii)$
$\angle BAL =\angle CAL \ldots \ldots . .( viii )$
In $\triangle BAL =\triangle CAL$
$ \angle BAL =\angle CAL \ldots[ [$From $(\text { viii }) ] $
$ \angle ABL =\angle ACL \ldots[$ From $(i) ] $
$ A B=A C \ldots[$ Given $]$
$ \therefore \triangle BAL \triangle CAL \ldots[ \text{ASA} ]$
$ \Rightarrow \angle A L B=\angle A L C \ldots\{\text { c. p.c.t ] }$
and $B L=L C \ldots \ldots . . .(i x) \ldots[\text { c. p.c.t ] }$
Now,
$\angle ALB +\angle ALC =180^{\circ}$
$ \Rightarrow \angle ALb +\angle ALB =180^{\circ}$
$ \Rightarrow 2 \angle ALB =180^{\circ}$
$ \Rightarrow \angle ALB =90^{\circ}$
$ \therefore AL \perp BC$
or $DL \perp BC$ and $BL = LC$
$\therefore DA$ produced bisects $BC$ at right angle.

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Question 25 Marks

In the given figure, $A B=A C$.

Prove that:$(i) DP = DQ;(ii) AP = AQ;(iii)AD$ bisects $\angle A$

Answer

Const: Join $AD.$
In $\triangle \mathrm{ABC}_r$
$A B=A C\dots.......[$Given$]$
$\therefore \angle \mathrm{C}=\angle \mathrm{B} \quad\dots......(i) [$angles opp. to equal sides are equal$]$
$(i)$
In $\triangle B P D$ and $\triangle C Q D$,
$\angle \mathrm{BPD}=\angle \mathrm{CQD} \ldots . . .[$ Each $=90^{\circ}]$
$ \angle \mathrm{B}=\angle \mathrm{C} \ldots . . . .[$ proved $]$
$ \mathrm{BD}=\mathrm{DC} \ldots . . . .[$ Given$]$
$ \therefore \triangle \mathrm{BPD} \cong \triangle \mathrm{CQD} \dots.......[\text{AAS}$ criterion$]$
$ \therefore \mathrm{DP}=\mathrm{DQ} \ldots . . .[\text { c.p.c.t}]$
$(ii)$
We have already proved that $\triangle \mathrm{BPD} \cong \triangle \mathrm{CQD}$ Therefore, $\mathrm{BP}=\mathrm{CQ}\dots.......[\text{c.p.c.t}]$
Now,
$A B=A C\dots.......[$Given$]$
$\Rightarrow A B-B P=A C-C Q$
$ \Rightarrow A P=A Q$
$(iii)$
In $\triangle \mathrm{APD}$ and $\triangle \mathrm{AQD}$,
$\mathrm{DP}=\mathrm{DQ} \ldots . . .[$proved$]$
$ \mathrm{AD}=\mathrm{AD} \ldots . . .[$ common$]$
$ \mathrm{AP}=\mathrm{AQ} \ldots . . .[$ Proved$]$
$ \therefore \triangle \mathrm{APD} \cong \triangle \mathrm{AQD} \ldots . .[\mathrm{SSS}]$
$ \Rightarrow \angle \mathrm{PAD}=\angle \mathrm{QAD} \ldots . . .[\text { c.p.c.t] }$
Hence,$ AD$ bisects $\angle A.$

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Question 35 Marks

The given figure shows an equilateral $\triangle ABC$ with each side $15\ cm$. Also, $DE \| BC , DF \| AC$, and $EG \| AB$.If $D E+D F+E G=20\ cm$, find $F G$.

Answer

$[\text{ABC}$ is an equilateral triangle.
Therefore, $AB = BC = AC = 15 \ cm$
$\angle A = \angle B = \angle C = 60^\circ In \triangle ADE, DE \| BC \dots........[$ Given $]$
$\angle AED = 60^\circ\dots ........[\because \angle ACB = 60^\circ ]$
$\angle ADE = 60^\circ \dots........[\because \angle ACB = 60^\circ ]$
$\angle DAE = 180^\circ − (60^\circ + 60^\circ ) = 60^\circ $
Similarly, $\text{BDF}$ and $\text{GEC}$ are equilateral triangles.
$= 60^\circ \dots.......[\because \angle C = 60^\circ ]$
Let $AD = x, AE = x, DE = x \dots......[\because \triangle ADE$ is an equilateral triangle$]$
Let $BD = y, FD = y, FB = y \dots......[\because \triangle BDF$ is an equilateral triangle$]$
Let $EC = z, GC = z , GE = z \dots...[\because \triangle GEC$ is an equilateral triangle$]$
Now,
$AD + DB = 15 \Rightarrow x + y = 15 \dots.......(i)$
$AE + EC = 15 \Rightarrow x + z = 15 \dots........(ii)$
Given, $DE + DF + EG = 20$
$\Rightarrow x + y + z = 20$
$\Rightarrow 15 + z = 20 \dots......[$From $(i)]$
$\Rightarrow z = 5$
From $(ii)$, we get $x = 10$
$\therefore y = 5$
Also, $BC = 15$
$BF + FG + GC = 15$
$\Rightarrow y + FG + z = 15$
$\Rightarrow 5 + FG + 5 = 15$
$\Rightarrow FG = 5$

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Question 45 Marks

If the equal sides of an isosceles triangle are produced, prove that the exterior angles so formed are obtuse and equal.

Answer

Const: $\mathrm{AB}$ is produced to $\mathrm{D}$ and $\mathrm{AC}$ is produced to $\mathrm{E}$
so that exterior angles $\angle \mathrm{DBC}$ and $\angle \mathrm{ECB}$ are formed.
In $\triangle \mathrm{ABC}$,
$A B=A C\dots........ [$ Given $]$
$\therefore \angle C=\angle B \ldots. (i) [$angels opp. to equal sides are equal$]$
Since angle $B$ and angle $C$ are acute they cannot be right angles or obtuse angles.
$ \angle \mathrm{ABC}+\angle \mathrm{DBC}=180^{\circ} \ldots . . .[\mathrm{ABD}$ is a st. line $]$
$ \angle \mathrm{DBC}=180^{\circ}-\angle \mathrm{ABC}$
$ \angle \mathrm{DBC}=180^{\circ}-\angle \mathrm{B} \ldots . . . \text { (ii) }$
Similarly,
$\angle \mathrm{ACB}+\angle \mathrm{ECB}=180^{\circ} \ldots . . .[\mathrm{ABD}$ is a st. line $]$
$ \angle \mathrm{ECB}=180^{\circ}-\angle \mathrm{ACB}$
$ \angle \mathrm{ECB}=180^{\circ}-\angle \mathrm{C} \ldots \ldots . . \text { (iii) }$
$ \angle \mathrm{ECB}=180^{\circ}-\angle \mathrm{B} \ldots . . . \text { (iv) }[$from $(i)$ and $(iii)]$
$ \angle \mathrm{DBC}=\angle \mathrm{ECB} \ldots . . .[$ from $(ii)$ and $(iv)]$
Now,
$\angle \mathrm{DBC}=180^{\circ}-\angle \mathrm{B}$
But $\angle B=$ Acute angel
$\therefore \angle \mathrm{DBC}=180^{\circ}-$ Acute angle $=$ obtuse angle
Similarly,
$\angle \mathrm{ECB}=180^{\circ}-\angle \mathrm{C} \text {. }$
But $\angle C=$ Acute angel
$\therefore \angle \mathrm{ECB}=180^{\circ}-$ Acute angle $=$ obtuse angle
Therefore, exterior angles formed are obtuse and equal.

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Question 55 Marks

Equal sides $AB$ and $AC$ of an isosceles $\triangle ABC$ are produced. The bisectors of the exterior angle so formed meet at $D$. Prove that $AD$ bisects $\angle A.$

Answer

$AB$ is produced to $E$ and $AC$ is produced to $F$.
$BD$ is the bisector of angle $C B E$ and $C D$ is the bisector of angle $BCF.$
$BD$ and $CD$ meet at $D$ In $\triangle A B C,AB = AC \ldots . . . .. [$Given$]$
$\therefore \angle C=\angle B [$angles opposite to equal sides are equal$]$
$\angle C B E=180^{\circ}-\angle B[\text{ABE}$ is a straight line$]$
$\Rightarrow \angle CBE =\frac{180^{\circ}-\angle B }{2} \ldots \ldots . .[ BD$ is bisector of $\angle CBE ]$
$\Rightarrow \angle CBE =90^{\circ}-\frac{\angle B }{2} \ldots \ldots \ldots .. (i)$
Similarly,
$\angle BCF =180^{\circ}-\angle C \ldots \ldots . .[\text{ACF}$ is a straight line $]$
$\Rightarrow \angle BCD =\frac{180^{\circ}-\angle C }{2} \ldots . . .[ CD$ is bisector of $\angle BCF]$
$\Rightarrow \angle BCD =90^{\circ}-\frac{\angle C }{2} \ldots \ldots .. (ii)$
Now,
$ \Rightarrow \angle CBD =90^{\circ}-\frac{\angle C }{2} \ldots \ldots .[\because \angle B =\angle C ]$
$ \Rightarrow \angle CBD =\angle BCD $
In $\triangle B C D$
$ \angle C B D=\angle B C D$
$ \therefore B D=C D$
In $\triangle A B D$ and $\triangle A C D$,
$AB = AC \ldots . . . .[$ Given$]$
$ AD = AD \ldots . . . .[$Common$]$
$ BD = CD \ldots . . . .[$ Proved$]$
$ \therefore \triangle ABD \cong \triangle ACD\dots....[\text {SSS}$ Criterion$]$
$ \Rightarrow \angle BAD =\angle CAD\dots ......[\text {c.p.c.t.}] $
Therefore, $A D$ bisects $\angle A$.

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Question 65 Marks

In an equilateral $\triangle ABC$; points $P, Q$ and $R$ are taken on the sides $AB, BC$ and $CA$ respectively such that $AP = BQ = CR$. Prove that $\triangle PQR$ is equilateral.

Answer

$AB = BC = CA \ldots . .. (i) [$Given$]$
$AP = BQ = CR \ldots . .. (ii) [$Given$]$
Subtracting $(ii)$ from $(i)$
$A B-A P=B C-B Q=C A-C R$
$BP = CQ = AR \ldots . . . .. (iii)$
$\therefore \angle A =\angle B =\angle C.......(iv) [$angles opp. to equal sides are equal$]$
In $\triangle BPQ$ and $\triangle CQR$
$BP = CQ \ldots . . .[$ From $(iii)]$
$ \angle B =\angle C \ldots . .[$ From $(iv)]$
$ BQ = CR \ldots . . .[$Given$]$
$ \therefore \triangle BPQ \cong \triangle CQR . . . . .[\text {SAS }$ criterion$]$
$ \Rightarrow PQ = QR \ldots . . . .( V )$
In $\triangle C Q R$ and $\triangle A P R$,
$CQ = AR \ldots . . . .[$ From $(iii)]$
$ \angle C =\angle A \ldots . . .[$ From $(iv)]$
$ CR = AP \ldots . . .[$Given$]$
$\therefore \triangle CQR \cong \triangle APR...[\text{SAS}$ criterion$]$
$\Rightarrow QR = PR \ldots( vi )$
From $(v)$ and $(vi)$
$P Q=Q R=P R$
Therefore, $\text{PQR}$ is an equilateral triangle.

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Question 75 Marks

Find $x$ :

Answer

Let us name the figure as following:

In $\triangle \mathrm{ABC}_r$
$\mathrm{AD}=\mathrm{AC} \ldots . . .[$ Given $]$
$\therefore \angle A D C=\angle A C D \quad......[$Angles opp. to equal sides are equal$]$
$\Rightarrow \angle \mathrm{ADC}=42^{\circ}$
Now,
$\angle \mathrm{ADC}=\angle \mathrm{DAB}+\angle \mathrm{DBA} \quad\dots... [$Exterior angle is equal to the sum of opp. interior angles$]$
But,
$\angle \mathrm{DAB}=\angle \mathrm{DBA}$
$[$Given: $\mathrm{BD}=\mathrm{DA}]$
$\therefore \angle \mathrm{ADC}=2 \angle \mathrm{DBA}$
$\Rightarrow 2 \angle \mathrm{DBA}=42^{\circ}$
$\Rightarrow \angle \mathrm{DBA}=21^{\circ}$
For $\mathrm{x}$ :
$\mathrm{x}=\angle \mathrm{CBA}+\angle \mathrm{BCA}[$Exterior angle is equal to the sum of opp. interior angles$]$
We know that,
$\angle C B A=21^{\circ}$
$\angle \mathrm{BCA}=42^{\circ}$
$\therefore x=21+42^{\circ}$
$\Rightarrow x=63^{\circ}$

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Question 85 Marks

An isosceles $\triangle A B C$ has $A C=B C. C D$ bisects $A B$ at $D$ and $\angle C A B=55^{\circ}$.Find:$(i) \angle DCB;(ii) \angle CBD.$

Answer

In $\triangle \mathrm{ABC}$,
$A C=B C\dots.......[$Given$]$
$\therefore \angle C A B=\angle C B D\dots........[$angles opp.to equal sides are equal$]$
$\Rightarrow \angle C B D=55^{\circ}$
In $\triangle \mathrm{ABC}$,
$ \angle \mathrm{CBA}+\angle \mathrm{CAB}+\angle \mathrm{ACB}=180^{\circ}$
but,$\angle \mathrm{CAB}=\angle \mathrm{CBA}=55^{\circ}$
$\Rightarrow 55^{\circ}+55^{\circ}+\angle \mathrm{ACB}=180^{\circ}$
$\Rightarrow \angle \mathrm{ACB}=180^{\circ}-110^{\circ}$
$\Rightarrow \angle \mathrm{ACB}=70^{\circ}$
Now,
In $\triangle A C D$ and $\triangle B C D$,
$ \mathrm{AC}=\mathrm{BC} \ldots . . .[$ Given$]$
$\mathrm{CD}=\mathrm{CD} \ldots . . . .[$ Common$]$
$\mathrm{AD}=\mathrm{BD} \ldots . . . .[$ Given: $\mathrm{CD}$ bisects $\mathrm{AB}]$
$\therefore \triangle \mathrm{ACD} \cong \triangle \mathrm{BCD}$
$\Rightarrow \angle \mathrm{DCA}=\angle \mathrm{DCB}$
$\Rightarrow \angle \mathrm{DCB}=\frac{\angle \mathrm{ACB}}{2}=\frac{70^{\circ}}{2}$
$\Rightarrow \angle \mathrm{DCB}=35^{\circ} .$

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Question 95 Marks

In the figure, given below, $A B=A C$.Prove that: $\angle B O C=\angle A C D$.

Answer

Let $\angle A B O=\angle O B C=x$ and $\angle A C O=\angle O C B=y$ In $\triangle \mathrm{ABC}$
$\angle B A C=180^{\circ}-2 x-2 y\ldots. (i)$
Since, $\angle B=\angle C\ldots[\mathrm{AB}=\mathrm{AC}]$
$\frac{1}{2} \mathrm{\sim B}=\frac{1}{2} \mathrm{C}$
$ \Rightarrow \mathrm{x}=\mathrm{y}$
Now,
$\angle \mathrm{ACD}=2 \mathrm{x}+\angle \mathrm{BAC} \ldots[$ Exterior angle is equal to sum of opp. interior angles $]$
$\angle \mathrm{ACD}=2 \mathrm{x}+180^{\circ}-2 \mathrm{x}-2 \mathrm{y} \ldots[$ From $(i) ]$
$\angle \mathrm{ACD}=180^{\circ}-2 \mathrm{y} \dots....(i)$
In $\triangle O B C_1$
$ \angle B O C=180^{\circ}-x-y$
$\Rightarrow \angle B O C=180^{\circ}-\mathrm{y}-\mathrm{y} \ldots[$ Already proved $]$
$\Rightarrow \angle \mathrm{BOC}=180^{\circ}-2 \mathrm{y} ...(ii)$
From $(i)$ and $(ii)$
$\angle B O C=\angle A C D$

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Question 105 Marks

Calculate $x$ :

Answer

Let the triangle be $\text{ABC}$ and the altitude be $A D$.

In $\triangle \mathrm{ABD}$, $\angle \mathrm{DBA}=\angle \mathrm{DAB}=37^{\circ} \ldots . .[$ Given $\mathrm{BD}=\mathrm{AD}$ and angles opposite to equal sides are equal$]$
Now, $\angle \mathrm{CDA}=\angle \mathrm{DBA}+\angle \mathrm{DAB} \ldots \ldots. [$Exterior angle is equal to the sum of opp. interior angles$]$
$\therefore \angle C D A=37^{\circ}+37^{\circ}$
$ \Rightarrow \angle C D A=74^{\circ}$
Now in $\triangle A D C$,
$\angle C D A=\angle C A D=74^{\circ} \ldots .[$ Given $\mathrm{CD}=\mathrm{AC}$ and angles opposite to equal sides are equal$]$
$ \angle C A D+\angle C D A+\angle A C D=180^{\circ}$
$ \Rightarrow 74^{\circ}+74^{\circ}+x=180^{\circ}$
$ \Rightarrow x=180^{\circ}-148^{\circ}$
$ \Rightarrow x=32^{\circ}$

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Question 115 Marks

In the following figure, $A B=A C ; B C=C D$ and $D E$ are parallel to $BC$.Calculate:$(i) \angle CDE;(ii) \angle DCE$

Answer

$\angle FAB = 128^\circ \dots.......[$ Given $]$
$\angle BAC + \angle FAB = 180^\circ \dots......[ \text{FAC}$ is a st. line $]$
$\Rightarrow \angle BAC = 180^\circ − 128^\circ $
$\Rightarrow \angle BAC = 52^\circ $ In $\triangle ABC,$
$\angle A = 52^\circ $
$\angle B= \angle C \dots.....[$Given $AB = AC$ and angels opposite to equal sides are equal$]$
$\angle A + \angle B + \angle C = 180^\circ $
$\Rightarrow \angle A + \angle B + \angle B = 180^\circ $
$\Rightarrow 52^\circ + 2\angle B = 180^\circ $
$\Rightarrow 2\angle B = 128^\circ $
$\Rightarrow \angle B = 64^\circ = \angle C ..\dots......(i)$
$\Rightarrow \angle B = \angle ADE \dots.......[$ Given $DE \| BC ]$
$(i)$
Now,
$\angle ADE + \angle CDE + \angle B = 180^\circ \dots....[ \text{ADB}$ is a st. line $]$
$\Rightarrow 64^\circ + \angle CDE + 64^\circ = 180^\circ $
$\Rightarrow \angle CDE = 180^\circ − 128^\circ $
$\Rightarrow \angle CDE = 52^\circ $
$(ii)$
Given $DE \| BC$ and $DC$ is the transversal.
$\Rightarrow \angle CDE = \angle DCB = 52^\circ\dots .......(ii)$
Also, $\angle ECB = 64^\circ \dots.....[$ From $(i) ]$
But,
$\angle ECB = \angle DCE + \angle DCB$
$\Rightarrow 64^\circ = \angle DCE + 52^\circ $
$\Rightarrow \angle DCE + 64^\circ − 52^\circ $
$\Rightarrow \angle DCE = 12^\circ $

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Question 125 Marks

Calculate:$(i) \angle ADC;(ii)\angle ABC;(iii) \angle B A C$

Answer

Given: $\text{ACE} = 130^\circ ; AD = BD = CD$
Proof:
$(i) \angle ACD + \angle ACE = 18^\circ\dots ....... [\text{DCE}$ is a st. line $]$
$\Rightarrow \angle ACD = 180^\circ − 130^\circ$
$\Rightarrow \angle ACD = 50^\circ$
Now, $CD = AD$
$\Rightarrow \angle ACD = \angle DAC = 50^\circ \dots..... (i)[$ Since angels opposite to equal sides are equal$]$
In $\triangle ADC,$
$\angle ACD = \angle DAC = 50^\circ$
$\angle ACD + \angle DAC + \angle ADC = 180^\circ$
$50^\circ + 50^\circ + \angle ADC = 180^\circ$
$\angle ADC =180^\circ − 100^\circ$
$\angle ADC = 80^\circ$
$(ii) \angle ADC = \angle ABD + \angle DAB \dots....[$Exterior angle is equal to sum of opp. interor angle$]$
But $AD = BD$
$\therefore \angle DAB = \angle ABD$
$\Rightarrow 80^\circ = \angle ABD + \angle ABD$
$\Rightarrow 2\angle BD = 80^\circ$
$\Rightarrow \angle ABD = 40^\circ = \angle DAB \dots....(ii)$
$(iii)\angle BAC = \angle DAB + \angle DAC$
substituting the value from $(i)$ and $(ii)$
$\angle BAC = 40^\circ + 50^\circ$
$\Rightarrow \angle BAC = 90^\circ$

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Question 135 Marks

In the given figure, $A D=A B=A C, B D$ is parallel to $C A$ and angle $\text{ACB}$
$=65^{\circ}$. Find $\angle DAC.$

Answer

We can see that the $\triangle ABC$ is an isosceles triangle with Side $AB =$ Side $AC.$
$\Rightarrow \angle ACB = \angle ABC$
As $\angle ACB = 65^\circ $
hence $\angle ABC = 65^\circ $
Sum of all the angles of a triangle is $180^\circ $
$\angle ACB + \angle CAB + \angle ABC = 180^\circ $
$65^\circ + 65^\circ + \angle CAB = 180^\circ $
$\angle CAB = 180^\circ − 130^\circ $
$\angle CAB = 50^\circ $ As $BD$ is parallel to $CA$
Therefore, $\angle CAB = \angle DBA$ since they are alternate angles.
$\angle CAB = \angle DBA = 50^\circ $
We see that $\triangle ADB$ is an isosceles triangle with Side $AD =$ Side $AB.$
$\Rightarrow \angle ADB = \angle DBA = 50^\circ $
Sum of all the angles of a triangle is $180^\circ $
$\angle ADB + \angle DAB + \angle DBA = 180^\circ $
$50^\circ + \angle DAB + 50^\circ = 180^\circ $
$\angle DAB = 180^\circ − 100^\circ = 80^\circ $
$\angle DAB = 80^\circ $
The $\angle DAC$ is the sum of $\angle DAB$ and $\text{CAB}.$
$\angle DAC = \angle CAB + \angle DAB$
$\angle DAC = 50^\circ + 80^\circ $
$\angle DAC = 130^\circ $

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Question 145 Marks

In $\triangle ABC, D$ is a point in $AB$ such that $AC = CD = DB$. If $\angle B = 28^\circ $, find the $\angle ACD.$

Answer

$\triangle \mathrm{DBC}$ is an isosceles triangle.
A s, side $C D=$ Side $DB$
$\Rightarrow \angle \mathrm{DBC}=\angle \mathrm{DCB} \ldots[$lf two sides of a triangle are equal, then angles opposite to them are equal$]$
And $\angle \mathrm{B}=\angle \mathrm{DBC}=\angle \mathrm{DCB}=28^{\circ}$
As the sum of all the angles of the triangle is $180^{\circ}$
$\angle \mathrm{DCB}+\angle \mathrm{DBC}+\angle \mathrm{BCD}=180^{\circ}$
$ \Rightarrow 28^{\circ}+28^{\circ}+\angle \mathrm{BCD}=180^{\circ}$
$ \Rightarrow \angle \mathrm{BCD}=180^{\circ}-56^{\circ}$
$ \Rightarrow \angle \mathrm{BCD}=124^{\circ}$
Sum of two non$-$adjacent interior angles of a triangle is equal to the exterior angle.
$\Rightarrow \angle \mathrm{DBC}+\angle \mathrm{DCB}=\angle \mathrm{ADC}$
$ \Rightarrow 28^{\circ}+28^{\circ}=\angle \mathrm{ADC}$
$ \Rightarrow \angle \mathrm{ADC}=56^{\circ}$
Now $\triangle \mathrm{ACD}$ is an isosceles triangle with $\mathrm{AC}=\mathrm{DC}$
$\Rightarrow \angle \mathrm{ADC}=\angle \mathrm{DAC}=56^{\circ}$
Sum of all the angles of a triangle is $180^{\circ}$
$\Rightarrow \angle \mathrm{ACD}+\angle \mathrm{ADC}+\angle \mathrm{DAC}=180^{\circ}$
$ \Rightarrow \angle \mathrm{ACD}+56^{\circ}+56^{\circ}=180^{\circ}$
$ \Rightarrow \angle \mathrm{ACD}=180^{\circ}-112^{\circ}$
$ \Rightarrow \angle \mathrm{ACD}=68^{\circ}$

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Question 155 Marks

$\text{ABC}$ is an equilateral triangle. Its side $BC$ is produced up to point $E$ such that $C$ is mid$-$point of $BE$. Calculate the measure of angles $\text{ACE}$ and $\text{AEC}.$

Answer

$\triangle \mathrm{ABC}$ is an equilateral triangle.
$\Rightarrow$ Side $A B=$ Side $A C$
$\Rightarrow \angle \mathrm{ABC}=\angle \mathrm{ACB}\dots........ [$If two sides of a triangle are equal, then angles opposite to them are equal$]$
Similarly , Side $A C=$ Side $B C$
$\Rightarrow \angle C A B=\angle A B C\dots....... [$If two sides of a triangle are equal, then angles opposite to them are equal$]$
Hence, $\angle A B C=\angle C A B=\angle A C B=y ($say$)$
As the sum of all the angles of the triangle is $180^{\circ}$.
$\angle \mathrm{ABC}+\angle \mathrm{CAB}+\angle \mathrm{ACB}=180^{\circ}$
$ \Rightarrow 3 \mathrm{y}=180^{\circ}$
$ \Rightarrow \mathrm{y}=60^{\circ}$
$\angle \mathrm{ACB}=\angle \mathrm{ACB}=\angle \mathrm{ABC}=60^{\circ}$
Sum of two non$-$adjacent interior angles of a triangle is equal to the exterior angle.
$\Rightarrow \angle \mathrm{CAB}+\angle \mathrm{CBA}=\angle \mathrm{ACE}$
$ \Rightarrow 60^{\circ}+60^{\circ}=\angle \mathrm{ACE}$
$ \Rightarrow \angle \mathrm{ACE}=120^{\circ}$
Now $\triangle \mathrm{ACE}$ is an isosceles triangle with $\mathrm{AC}=\mathrm{CF}$
$\Rightarrow \angle \mathrm{EAC}=\angle \mathrm{AEC}$
Sum of all the angles of a triangle is $180^{\circ}$
$\angle \mathrm{EAC}+\angle \mathrm{AEC}+\angle \mathrm{ACE}=180^{\circ}$
$ \Rightarrow 2 \angle \mathrm{AEC}+120^{\circ}=180^{\circ}$
$ \Rightarrow 2 \angle \mathrm{AEC}=180^{\circ}-120^{\circ}$
$ \Rightarrow \angle \mathrm{AEC}=30^{\circ}$

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Question 165 Marks

In the given figure; $A E\|B D, A C\| E D$ and $A B=A C$. Find $\angle a , \angle b$ and $\angle C$.

Answer

Let $\mathrm{P}$ and $\mathrm{Q}$ be the points as shown below:

Given:
$\angle \mathrm{PDQ}=58^{\circ}$
$ \angle \mathrm{PDQ}=\angle \mathrm{EDC}=58^{\circ} \ldots . .[$ Vertically opp . angles $]$
$ \angle \mathrm{EDC}=\angle \mathrm{ACB}=58^{\circ} \ldots . . .[$ Corresponding angles $\because \mathrm{AC} \| \mathrm{ED}]$
In $\triangle A B C$,
$\mathrm{AB}=\mathrm{AC}\dots.......[$ Given $]$
$\therefore \angle \mathrm{ACB}=\angle \mathrm{ABC}=58^{\circ}\dots......[$angels opp. to equal sides are equal$]$
Now,
$\angle \mathrm{ACB}+\angle \mathrm{ABC}+\angle \mathrm{BAC}=180^{\circ}$
$ \Rightarrow 58^{\circ}+58^{\circ}+\mathrm{a}=180^{\circ}$
$ \Rightarrow \mathrm{a}=180^{\circ}-116^{\circ}$
$ \Rightarrow \mathrm{a}=64^{\circ}$
Since $A E \| B D$ and $A C$ is the transversal.
$\angle \mathrm{ABC}=\mathrm{b} \ldots . . .[$ Corresponding angles $]$
$\therefore \mathrm{b}=58^{\circ}$
Also since $\mathrm{AE}\| \mathrm{BD}$ and $\mathrm{ED}$ is the transversal $\angle \mathrm{EDC}=\mathrm{C}\dots....... $[ Corresponding angles $]$
$\therefore \mathrm{C}=58^{\circ}$

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[5 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip